Trigonometric Ratios of Sub-Multiple Angle

 


Trigonometric Ratios of Sub-Multiple Angle  

Formula of Sub-multiple Angles 


\begin{align} (1) & \sin A = 2\sin \frac{A}{2} \cos \frac{A}{2} \\ (2) & \sin A = \frac{2\tan \frac{A}{2} }{1+\tan^2 \frac{A}{2} } \\ (3) & \sin A = \frac{2\cot \frac{A}{2} }{1+\cot^2 \frac{A}{2} } \\ (4) & \cos  A = \cos^2 \frac{A}{2} -\sin^2 \frac{A}{2}  \\ (5) & \cos A = 2\cos^2 \frac{A}{2} - 1 \\ (6) & \cos  A =1 - 2\sin^2 \frac{A}{2} \\ (7) & \cos A = \frac{1-\tan^2 \frac{A}{2}}{1+\tan^2 \frac{A}{2}} \\ (8) & \cos A =\frac{\cos^2 \frac{A}{2} -1}{\cot^2 \frac{A}{2} + 1} \\ (9) & \tan A = \frac{2\tan \frac{A}{2} }{1-\tan^2 \frac{A}{2}}\\ (10) & \cot A = \frac{\cot^2 \frac{A}{2} -1}{2\cot \frac{A}{2} } \\ (11) & \sin A = 3\sin \frac{A}{3} - 4\sin^3 \frac{A}{3}\\ (12) & \cos A = 4\cos^3 \frac{A}{3} - 3\cos \frac{A}{3}  \\ (13) & \tan A = \frac{3\tan \frac{A}{3} - \tan^3 \frac{A}{3}}{1- 3\tan^2 \frac{A}{3}} \\ (14) & \cot A = \frac{3\cot \frac{A}{3} - \cot^3 \frac{A}{3} }{1-3\cot^2 \frac{A}{3}} \end{align} 

1. (a) Write down the formula of \( \sin A \) in terms of \( \tan \frac{A}{2} .\)
Solution: \begin{align} \sin A = \frac{2\tan \frac{A}{2} }{1+\tan^2 \frac{A}{2} } \end{align}


1b Write down the formula of \( \sin \theta \) in terms of \( \cot \frac{\theta}{2} .\)
Solution:  \begin{align} \sin \theta = \frac{2\cot \frac{\theta}{2} }{1+\cot^2 \frac{\theta}{2} } \end{align}


1c Write down the formula of \( \cos A \) in terms of : \\ (i) \( \sin \frac{A}{2} \)  (ii) \( \cos \frac{A}{2} \)  (iii) \( \tan \frac{A}{2} \)  (iv) \( \cot \frac{A}{2} \).
Solution: \begin{align} & \text{(i)}  \cos  A =1 - 2\sin^2 \frac{A}{2} \\ &  \text{(ii)} \cos A = 2\cos^2 \frac{A}{2} - 1 \\ &  \text{(iii)}  \cos A = \frac{1-\tan^2 \frac{A}{2}}{1+\tan^2 \frac{A}{2}} \\ &  \text{(iv)}  \cos A =\frac{\cos^2 \frac{A}{2} -1}{\cot^2 \frac{A}{2} + 1} \\ \end{align}

2 (a) Find the value of \( \sin A , \cos A , \tan A \text{ and } \cot A \) when, \( \sin \frac{A}{2} = \frac{3}{5} \)
Solution: \begin{align} \sin \frac{A}{2} & = \frac{3}{5} \\ \cos \frac{A}{2}  & = \sqrt{1-\sin^2 \frac{A}{2}} \\ & = \sqrt{1-\left(\frac{3}{5} \right)^2} \\ & = \sqrt{ 1 - \frac{9}{25} } \\ & = \sqrt{\frac{25 - 9 }{25}} \\ & = \sqrt{\frac{16}{25}}\\ & =\frac{4}{5} \\ \therefore \sin \frac{A}{2} & =\frac{4}{5} \\ \text{Now, } \sin A & = 2\sin \frac{A}{2} \cos \frac{A}{2} \\ & = 2\times \frac{3}{5} \times \frac{4}{5} \\ & = \frac{24}{25} \\ \therefore \sin A & = \frac{24}{25} \end{align} \begin{align} \text{Also} \\ \cos A & = 1 - 2\sin^2 \frac{A}{2} \\ & = 1 - 2 \left( \frac{3}{5}  \right)^2\\ & = 1 - 2 \left(\frac{9}{25} \right)\\ & = \frac{25 - 18}{25} \\ & = \frac{7}{25} \\ \therefore  \cos A & = \frac{7}{25} \end{align} \begin{align} \text{Also} \\ \tan A & = \frac{\sin A}{\cos A} \\ & = \frac{\dfrac{24}{25}}{\dfrac{7}{25}}\\ \therefore  \tan A & = \frac{24}{7} \end{align} \begin{align} \text{Also} \\ \cot A & = \frac{1}{\tan A} \\ & = \frac{1}{\dfrac{24}{7}}\\ & = \frac{7}{24} \\ \therefore \cot A & = \frac{7}{24} \end{align}


2. (b) Find the value of \( \sin A , \cos A , \tan A \text{ and } \cot A \) when, \(  \cos \frac{A}{2} = \frac{4}{5} \)
Solution:  \begin{align} \cos \frac{A}{2} & = \frac{4}{5} \\ \sin \frac{A}{2} & = \sqrt{1-\cos^2 \frac{A}{2}}\\ & = \sqrt{1-\cos^2 \frac{A}{2}}\\ & = \sqrt{1 - \left( \frac{4}{5} \right)^2} \\ & = \sqrt{1 - \frac{16}{25} }\\ & = \sqrt{\frac{25 - 16}{25} }\\ & = \sqrt{\frac{9}{25} } \\ & = \frac{3}{5} \\ \therefore \sin \frac{A}{2} & = \frac{3}{5}\\ \text{Now, } \sin A & = 2\sin \frac{A}{2} \cos \frac{A}{2} \\ & = 2\times \frac{3}{5} \times \frac{4}{5} \\ & = \frac{24}{25} \\ \therefore \sin A & = \frac{24}{25} \end{align} \begin{align} \text{Also} \\ \cos A & = 1 - 2\sin^2 \frac{A}{2} \\ & = 1 - 2 \left( \frac{3}{5}  \right)^2\\ & = 1 - 2 \left(\frac{9}{25} \right)\\ & = \frac{25 - 18}{25} \\ & = \frac{7}{25} \\ \therefore  \cos A & = \frac{7}{25} \end{align} \begin{align} \text{Also} \\ \tan A & = \frac{\sin A}{\cos A} \\ & = \frac{\dfrac{24}{25}}{\dfrac{7}{25}}\\ \therefore  \tan A & = \frac{24}{7} \end{align} \begin{align} \text{Also} \\ \cot A & = \frac{1}{\tan A} \\ & = \frac{1}{\dfrac{24}{7}}\\ & = \frac{7}{24} \\ \therefore \cot A & = \frac{7}{24} \end{align}


2. (c) Find the value of \( \sin A , \cos A , \tan A \text{ and } \cot A \) when, \( \tan \frac{A}{2} = \frac{5}{12} \)
Solution:  \begin{align} \tan \frac{A}{2} & = \frac{5}{12} \\ \sin A & = \frac{2\tan  \frac{A}{2}}{1+\tan^2  \frac{A}{2} }\\ & = \frac{2\times \dfrac{5}{12}}{1+\left( \dfrac{5}{12} \right)^2 }\\ & = \frac{\dfrac{5}{6}}{1+\dfrac{25}{144}} \\ & = \frac{\dfrac{5}{6}}{\dfrac{144+25}{144}}\\ & = \frac{\dfrac{5}{6}}{\dfrac{169}{144}}\\ & = \frac{5}{6} \times \frac{144}{169} \\ & = \frac{120}{169} \end{align} \begin{align} \tan \frac{A}{2} & = \frac{5}{12} \\ \cos A & = \frac{1-\tan^2 \frac{A}{2} }{1+\tan^2 \frac{A}{2} }\\ \cos A & = \frac{1-\left(\dfrac{5}{12}\right)^2 }{1-\left(\dfrac{5}{12}\right)^2}\\ & = \frac{1- \dfrac{25}{144} }{1+ \dfrac{25}{144}}\\ & = \frac{\dfrac{144 -25}{144} }{\dfrac{144+25}{144}}\\ & = \frac{ 119}{169} \end{align} \begin{align} \tan \frac{A}{2} & = \frac{5}{12} \\ \tan A & = \frac{2\tan  \frac{A}{2}}{1-\tan^2  \frac{A}{2} }\\ & = \frac{2\times \dfrac{5}{12}}{1-\left( \dfrac{5}{12} \right)^2 }\\ & = \frac{\dfrac{5}{6}}{1-\dfrac{25}{144}} \\ & = \frac{\dfrac{5}{6}}{\dfrac{144-25}{144}}\\ & = \frac{\dfrac{5}{6}}{\dfrac{119}{144}}\\ & = \frac{5}{6} \times \frac{144}{119} \\ & = \frac{120}{119} \end{align} \begin{align} \text{Also} \\ \cot A & = \frac{1}{\tan A} \\ & = \frac{1}{\dfrac{120}{119}}\\ & = \frac{119}{120} \\ \therefore \cot A & = \frac{119}{120} \end{align}


2. (d) Find the value of \( \sin A , \cos A , \tan A \text{ and } \cot A \) when, \( \cot \frac{A}{2} = \frac{3}{4} \)
Solution:  \begin{align} \cot \frac{A}{2} & = \frac{3}{4}\\ \tan \frac{A}{2} & = \frac{4}{3} \\ \sin A & = \frac{2\tan  \frac{A}{2}}{1+\tan^2  \frac{A}{2} }\\ & = \frac{2\times \dfrac{4}{3}}{1+\left( \dfrac{4}{3} \right)^2 }\\ & = \frac{\dfrac{8}{3}}{1+\dfrac{16}{9}} \\ & = \frac{\dfrac{8}{3}}{\dfrac{9+16}{9}}\\ & = \frac{\dfrac{8}{3}}{\dfrac{25}{9}}\\ & = \frac{8}{3} \times \frac{9}{25} \\ & = \frac{24}{25} \end{align} \begin{align} \tan \frac{A}{2} & = \frac{4}{3} \\ \cos A & = \frac{1-\tan^2 \frac{A}{2} }{1+\tan^2 \frac{A}{2} }\\ \cos A & = \frac{1-\left(\dfrac{4}{3}\right)^2 }{1-\left(\dfrac{4}{3}\right)^2}\\ & = \frac{1- \dfrac{16}{9} }{1+ \dfrac{16}{9}}\\ & = \frac{\dfrac{9 -16}{9} }{\dfrac{9+16}{9}}\\ & = \frac{ -7}{25} \end{align} \begin{align} \tan \frac{A}{2} & = \frac{4}{3} \\ \tan A & = \frac{2\tan  \frac{A}{2}}{1-\tan^2  \frac{A}{2} }\\ & = \frac{2\times \dfrac{4}{3}}{1-\left( \dfrac{4}{3} \right)^2 }\\ & = \frac{\dfrac{8}{3}}{1-\dfrac{16}{9}} \\ & = \frac{\dfrac{8}{3}}{\dfrac{9-16}{9}}\\ & = \frac{\dfrac{8}{3}}{\dfrac{-7}{9}}\\ & = \frac{8}{3} \times \frac{9}{-7} \\ & = \frac{24}{-7}\\ & = -\frac{24}{7} \end{align} \begin{align} \text{Also} \\ \cot A & = \frac{1}{\tan A} \\ & = \frac{1}{-\dfrac{24}{7}}\\ & = -\frac{7}{24} \\ \therefore \cot A & = -\frac{7}{24} \end{align}

Find the values of half angle under the given condition.

3. (a) If \( \cos 30^\circ = \frac{\sqrt{3}}{2} \), find the value of \( \cos 15^\circ \).
Solution:  \begin{align} \text{ Given }, \cos 30^\circ & = \frac{\sqrt{3}}{2} & \\ \text{ We know },\ & \\ \sin^2 A \ & =\frac{1-\cos2A}{2}\\ \sin^2 15^{\circ} \ &  = \frac{1-\cos30^{\circ}}{2}\\ \ \ &  =  \frac{1-\frac{\sqrt{3}}{2}}{2}\\ \ \ &  = \frac{\frac{2-\sqrt{3}}{2}}{2}\\ \ \ &  = \frac{2-\sqrt{3}}{4}\\ \ \ &  = \frac{2-\sqrt{3}}{4}\times \frac{2}{2}\\ \ \ &  = \frac{ 4 -2 \sqrt{3}}{8}\\ \ \ &  = \frac{ 3 -2 \sqrt{3}+1}{8}\\ \ \ &  = \frac{ (\sqrt{3})^2 -2\times  \sqrt{3}\times 1+1^2}{(2\sqrt{2})^2}\\ \ \ &   = \frac{(\sqrt{3}-1)^2}{(2\sqrt{2})^2}\\ \ \ &  = \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2\\ \text{or,} \sin^2 15^{\circ} \ &  = \left(\frac{\sqrt{3}-1}{2\sqrt{2}} \right)^2\\ \therefore \sin 15^{\circ} \ &  = \frac{\sqrt{3}-1}{2\sqrt{2}} \\ \end{align}

3. (b) If \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), find the value of \( \cos 22 \frac{1}{2}^\circ \).
Solution: \begin{align} \text{ Given },\cos 45^\circ & = \frac{1}{\sqrt{2}} \\ \text{We know,}\ & \\ \cos^2 A\ & =\frac{1+\cos2A}{2}\\ \text{or, }\cos^2 22\frac{1}{2}^{\circ}\ & =\frac{1+\cos45^{\circ}}{2}\\  \ &  = \frac{1+\dfrac{1}{\sqrt{2}}}{2}\\\ &  = \frac{\dfrac{\sqrt{2}+1}{\sqrt{2}}}{2}\\  \ &  = \frac{\sqrt{2}+1}{2\sqrt{2}}\\  \ &  = \frac{\sqrt{2}+1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ \ &  = \frac{2+\sqrt{2}}{4}\\ \text{or, }\cos^2 22\frac{1}{2}^{\circ}\ & = \frac{2+\sqrt{2}}{4}\\ \text{or, }\cos^2 22\frac{1}{2}^{\circ}\ & =\sqrt{\frac{2+\sqrt{2}}{4}}\\ \therefore \cos 22\frac{1}{2}^{\circ}\ & =\frac{1}{2}\sqrt{2+\sqrt{2}}\\ \end{align}

3.(c) If \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), find the value of \( \tan 22 \frac{1}{2}^\circ \). 
Solution:  \begin{align}  \text{We know,}\ & \\\sin 45^\circ & = \frac{1}{\sqrt{2}}  \\ \cos 45^\circ & = \sqrt{1-\sin^2 45^\circ }\\ & = \sqrt{ 1- \left(\frac{1}{\sqrt{2}}\right)^2} \\ & = \sqrt{ 1 - \frac{1}{2}} \\ & = \sqrt{\frac{2-1}{2}} \\ & = \sqrt{ \frac{1}{2}}\\ & = \frac{1}{\sqrt{2}} \\ \therefore \cos 45^\circ = & = \frac{1}{\sqrt{2}}  \end{align} \begin{align}  \text{We know,}\ & \\ \tan^2 A\ & =\frac{1-\cos2A}{1+\cos2A}\\ \text{or, }\tan^2 22\frac{1}{2}^{\circ}\ & =\frac{1-\cos45^{\circ}}{1+\cos45^{\circ}}\\  \ &  = \frac{1-\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}\\  \ &  = \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{\frac{\sqrt{2}+1}{\sqrt{2}}}\\  \ &  = \frac{\sqrt{2}-1}{\sqrt{2}+1}\\   \ &  =  \frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}\\  \ &  = \frac{\left(\sqrt{2}-1\right)^2}{(\sqrt{2})^2-1^2}\\   \ &  = \frac{(\sqrt{2})^2 -2\times \sqrt{2}\times 1 +1^2 }{2-1}\\  \ &  = \frac{2-2\sqrt{2}+1}{1}\\ \ &  = 3-2\sqrt{2}\\ \text{or, }\tan^2 22\frac{1}{2}^{\circ}\ & =  3-2\sqrt{2}\\ \therefore \tan 22\frac{1}{2}^{\circ}\ & = \sqrt{ 3-2\sqrt{2}}\\ \end{align}

Prove under the given condition.

4. (a) If \( \cos 30^\circ = \frac{\sqrt{3}}{2} \), show that \( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
Solution:  \begin{align} \text{ Given }, \cos 30^\circ & = \frac{\sqrt{3}}{2} & \\ \text{ We know },\ & \\ \sin^2 A \ & =\frac{1-\cos2A}{2}\\ \sin^2 15^{\circ} \ &  = \frac{1-\cos30^{\circ}}{2}\\ \ \ &  =  \frac{1-\frac{\sqrt{3}}{2}}{2}\\ \ \ &  = \frac{\frac{2-\sqrt{3}}{2}}{2}\\ \ \ &  = \frac{2-\sqrt{3}}{4}\\ \ \ &  = \frac{2-\sqrt{3}}{4}\times \frac{2}{2}\\ \ \ &  = \frac{ 4 -2 \sqrt{3}}{8}\\ \ \ &  = \frac{ 3 -2 \sqrt{3}+1}{8}\\ \ \ &  = \frac{ (\sqrt{3})^2 -2\times  \sqrt{3}\times 1+1^2}{(2\sqrt{2})^2}\\ \ \ &   = \frac{(\sqrt{3}-1)^2}{(2\sqrt{2})^2}\\ \ \ &  = \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2\\ \text{or,} \sin^2 15^{\circ} \ &  = \left(\frac{\sqrt{3}-1}{2\sqrt{2}} \right)^2\\ \therefore \sin 15^{\circ} \ &  = \frac{\sqrt{3}-1}{2\sqrt{2}} \\ \end{align}

4. (b) If \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), show that \( \sin 22 \frac{1}{2}^\circ = \frac{1}{2} \sqrt{2-\sqrt{2}}  \)
Solution:  \begin{align}   \text{We know,}\ & \\ \sin^2 A\ & =\frac{1-\cos2A}{2}\\ \text{or, }\sin^2 22\frac{1}{2}^{\circ}\ & =\frac{1-\cos45^{\circ}}{2}\\  \ &  = \frac{1-\frac{1}{\sqrt{2}}}{2}\\  \ &  = \frac{\frac{\sqrt{2}-1}{\sqrt{2}}}{2}\\  \ &  = \frac{\sqrt{2}-1}{2\sqrt{2}}\\  \ &  = \frac{\sqrt{2}-1}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\\ &  = \frac{2-\sqrt{2}}{4}\\ \text{or, }\sin^2 22\frac{1}{2}^{\circ}\ & = \frac{2-\sqrt{2}}{4}\\ \text{or, }\sin^2 22\frac{1}{2}^{\circ}\ & =\sqrt{\frac{2-\sqrt{2}}{4}}\\ \therefore \sin 22\frac{1}{2}^{\circ}\ & =\frac{1}{2}\sqrt{2-\sqrt{2}}\\ \end{align}

4. (c) If \( \cos 330^\circ = \frac{\sqrt{3}}{2} \), show that \( \sin 165^\circ =  \frac{\sqrt{3} - 1}{2\sqrt{2}} \).
Solution:  \begin{align} \text{ Given }, \cos 330^\circ & = \frac{\sqrt{3}}{2} & \\ \text{ We know },\ & \\ \sin^2 A \ & =\frac{1-\cos2A}{2}\\ \sin^2 165^{\circ} \ &  = \frac{1-\cos330^{\circ}}{2}\\ \ \ &  =  \frac{1-\dfrac{\sqrt{3}}{2}}{2}\\ \ \ &  = \frac{\dfrac{2-\sqrt{3}}{2}}{2}\\ \ \ &  = \frac{2-\sqrt{3}}{4}\\ \ \ &  = \frac{2-\sqrt{3}}{4}\times \frac{2}{2}\\ \ \ &  = \frac{ 4 -2 \sqrt{3}}{8}\\ \ \ &  = \frac{ 3 -2 \sqrt{3}+1}{8}\\ \ \ &  = \frac{ (\sqrt{3})^2 -2\times  \sqrt{3}\times 1+1^2}{(2\sqrt{2})^2}\\ \ \ &   = \frac{(\sqrt{3}-1)^2}{(2\sqrt{2})^2}\\ \ \ &  = \left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)^2\\ \text{or,} \sin^2 165^{\circ} \ &  = \left(\frac{\sqrt{3}-1}{2\sqrt{2}} \right)^2\\ \therefore \sin 165^{\circ} \ &  = \frac{\sqrt{3}-1}{2\sqrt{2}} \\ \end{align}

4. (d) If \( \cos 30^\circ = \frac{\sqrt{3}}{2}\), show that: \( \tan 15^\circ = 2 - \sqrt{3} \)
Solution:  \begin{align} (iii)  \text{ We know },\ & \\ \tan^2 A \ & =\frac{1-\cos2A}{1+\cos2A}\\ \tan^2 15^{\circ} \ &  = \frac{1-\cos30^{\circ}}{1+\cos30^{\circ}}\\ \ \ &  =  \frac{1-\dfrac{\sqrt{3}}{2}}{1+\dfrac{\sqrt{3}}{2}}\\ \ \ &  = \frac{\dfrac{2-\sqrt{3}}{2}}{\dfrac{2+\sqrt{3}}{2}}\\ \ \ &  = \frac{2-\sqrt{3}}{2+\sqrt{3}}\\ \ \ &  = \frac{2-\sqrt{3}}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}\\ \ \ &  = \frac{\left(2-\sqrt{3}\right)^2}{4-3}\\ \ \ &  = \frac{(2-\sqrt{3})^2}{1}\\ \ \ &  = (2-\sqrt{3})^2 \\ \text{or,} \tan^2 15^{\circ} \ &  = (2-\sqrt{3})^2 \\ \therefore \tan 15^{\circ} \ &  = 2-\sqrt{3}\\ \end{align}

5. (a) If \( \cos \frac{\theta}{2} = \frac{1}{2} \left( a + \frac{1}{a} \right)\), show that: \( \cos \theta = \frac{1}{2} \left(a^2 + \frac{1}{a^2} \right) \)
Solution: \begin{align} \text{Given,}& \\  \cos \frac{\theta}{2} & = \frac{1}{2} \left( a + \frac{1}{a} \right) \\  \text{Now, }\\\text{LHS }   \ \ &  =\cos\theta \\ \ \ &  =2\cos^2\frac{\theta}{2}-1 \\    \ \ &  =2\left(\cos\frac{\theta}{2}\right)^2-1 \\ \ \ &  =2\left\{\frac{1}{2}\left(a+\frac{1}{a}\right)\right\}^2-1 \\ \ \ &  =2\times\frac{1}{4}\left(a+\frac{1}{a}\right)^2-1 \\ \ \ &  =\frac{1}{2}\left(a+\frac{1}{a}\right)^2-1 \\ \ \ &  =\frac{1}{2}\left(a^2+2\times a\times\frac{1}{a}+\frac{1}{a^2}\right)-1 \\  \ \ &  =\frac{1}{2}\left(a^2+\frac{1}{a^2}+2\right)-1 \\ \ \ &  =\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)+\frac{1}{2}\times2-1 \\ \ \ &  =\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)+1-1 \\ \ \ &  =\frac{1}{2}\left(a^2+\frac{1}{a^2}\right) \\ \ \ &  = \text{ RHS} \end{align}

5. (b) If \( \sin \frac{\theta}{2} = \frac{1}{2} \left( m + \frac{1}{m} \right)\), show that: \( \cos \theta = - \frac{1}{2} \left(m^2 + \frac{1}{m^2} \right) \)
Solution:  \begin{align} \text{LHS } \ &  =\cos\theta\\ \ &  =1-2\sin^2\frac{\theta}{2}\\ \ &  =1-2\left(\sin\frac{\theta}{2}\right)^{^2}\\ \ &  =1-2\left\{\frac{1}{2}\left(m+\frac{1}{m}\right)\right\}^2\\ \ &  =1-2\times\frac{1}{4}\left(m+\frac{1}{m}\right)^2\\  \ &  =1-\frac{1}{2}\left(m^2+2\times m\times\frac{1}{m}+\frac{1}{m^2}\right)\\ \ &  =1-\frac{1}{2}\left(m^2+2+\frac{1}{m^2}\right)\\ \ &  =1-\frac{1}{2}\left(m^2+\frac{1}{m^2}+2\right)\\ \ &  =1-\frac{1}{2}\left(m^2+\frac{1}{m^2}\right)-\frac{1}{2}\times2\\ \ &  =1-\frac{1}{2}\left(m^2+\frac{1}{m^2}\right)-1\\ \ &  =-\frac{1}{2}\left(m^2+\frac{1}{m^2}\right)\\ \ &  =\text{ RHS } \end{align} 

 5. (c) If \( \cos\theta=\frac{1}{2}\left(p+\frac{1}{p}\right)\), then prove that: \(\cos3\theta=\frac{1}{2}\left(p^3+\frac{1}{p^3}\right).\) 
Solution:  \begin{align} \text{LHS} &=\cos\theta\\ &=4\cos^3\frac{\theta}{3} -\cos \frac{\theta}{3}\\ &=4\left(\cos\frac{\theta}{3}\right)^3-3\cos\frac{\theta}{3}\\ &=4\left\{\frac{1}{2}\left(p+\frac{1}{p}\right)^3\right\}-3\times\frac{1}{2}\left(p+\frac{1}{p}\right)^2\\ &=4\times\frac{1}{8}\left(p +\frac{1}{p}\right)^3-\frac{3}{2}\left( p +\frac{1}{p}\right)\\ &=\frac{1}{2}\left( p +\frac{1}{p}\right)^3-\frac{3}{2}\left(p+\frac{1}{p}\right)\\ &=\frac{1}{2}\left\{p^3+\frac{1}{p^3}+ 3\times p \times \frac{1}{p}\left(p +\frac{1}{p} \right) \right\}-\frac{3}{2}\left(p +\frac{1}{p}\right)\\ &=\frac{1}{2}\left\{p^3+\frac{1}{p^3}+ 3\left(p+\frac{1}{p} \right) \right\}-\frac{3}{2}\left(p+\frac{1}{p}\right)\\ & = \frac{1}{2} \left( p^3 + \frac{1}{p^3} \right) +\frac{3}{2} \left( p + \frac{1}{p} \right) - \frac{3}{2} \left( p + \frac{1}{p} \right)\\ & = \frac{1}{2} \left( p^3 + \frac{1}{p^3} \right)\\ &=\text{RHS} \end{align}
Alternative \begin{align} \text{LHS} &=\cos \theta\\ &=4\cos^3\frac{\theta}{3}-3\cos\frac{\theta}{3}\\ &=4\left(\cos\frac{\theta}{3}\right)^3-3\cos\frac{\theta}{3}\\ &=4\left\{\frac{1}{2}\left(p +\frac{1}{p}\right)^3\right\}-3\times\frac{1}{2}\left( p +\frac{1}{p}\right)^2\\ &=4\times\frac{1}{8}\left(p +\frac{1}{p}\right)^3-\frac{3}{2}\left(a+\frac{1}{p}\right)\\ &=\frac{1}{2}\left( p +\frac{1}{p}\right)^3-\frac{3}{2}\left(p+\frac{1}{p}\right)\\ &=\frac{1}{2}\left\{\left(p +\frac{1}{p}\right)^3-3\left(p +\frac{1}{p}\right)\right\}\\ &=\frac{1}{2}\left\{\left(p +\frac{1}{p}\right)^3-3\times p \times\frac{1}{p}\left( p +\frac{1}{p}\right)\right\}\\ &=\frac{1}{2}\left(p^3+\frac{1}{p^3}\right)\\    &\left[\because x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\right]\\ &=\text{RHS} \end{align}


5. (d) If \(\sin\frac{\theta}{3}=\frac{1}{2}\left(x+\frac{1}{x}\right)\) , show that \( \sin\theta=-\frac{1}{2}\left(x^3+\frac{1}{x^3}\right)\)
Solution:  \begin{align} \text{LHS } \ & =\sin\theta\\ \ & =3\sin\frac{\theta}{3}-4\sin^3\frac{\theta}{3}\\  \ & =3\times\frac{1}{2}\left(x+\frac{1}{x}\right)-4\times\left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}^3\\   \ & =\frac{3}{2}\left(x+\frac{1}{x}\right)-4\times\frac{1}{8}\left(x+\frac{1}{x}\right)^3\\ \ & =\frac{3}{2} \left( x + \frac{1}{x} \right) - \frac{1}{2} \left(x+\frac{1}{x}\right)^3\\ \ & =\frac{3}{2} \left( x + \frac{1}{x} \right) - \frac{1}{2} \left\{x^3+  \frac{1}{x^3}+3\times x\times \frac{1}{x} \left( x+ \frac{1}{x} \right) \right\}\\ \ & =\frac{3}{2} \left( x + \frac{1}{x} \right) - \frac{1}{2} \left\{x^3+  \frac{1}{x^3}+3 \left( x + \frac{1}{x} \right) \right\}\\ \ & =\frac{3}{2} \left( x + \frac{1}{x} \right) - \frac{1}{2} \left(x^3+  \frac{1}{x^3}\right) - \frac{3}{2} \left( x+ \frac{1}{x} \right) \\ \ & = - \frac{1}{2} \left(x^3+  \frac{1}{x^3}\right) \\ \ & =\text{ RHS} \end{align}
Alternative \begin{align} \text{LHS } \ & =\sin\theta\\ \ & =3\sin\frac{\theta}{3}-4\sin^3\frac{\theta}{3}\\  \ & =3\times\frac{1}{2}\left(x+\frac{1}{x}\right)-4\times\left\{\frac{1}{2}\left(x+\frac{1}{x}\right)\right\}^3\\   \ & =\frac{3}{2}\left(x+\frac{1}{x}\right)-4\times\frac{1}{8}\left(x+\frac{1}{x}\right)^3\\   \ & =\frac{1}{2}\left\{3\left(x+\frac{1}{x}\right)-\left(x+\frac{1}{x}\right)^3\right\}\\     \ & =-\frac{1}{2}\left\{\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)\right\}\\   \ & =-\frac{1}{2}\left\{\left(x+\frac{1}{x}\right)^3-3\times x\times\frac{1}{x}\left(a+\frac{1}{x}\right)\right\}\\ \ & =-\frac{1}{2}\left(x^3+\frac{1}{x^3}\right) \left[\because a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\right]\\ \ & =\text{ RHS}\\ \end{align}

 6. (a) Prove that: \( \cos \frac{\theta}{2} = \pm \sqrt{\frac{1+\cos \theta}{2}} \)
Solution:  First Method \begin{align} \text{We know, }\\\cos \theta &= 2\cos^2 \frac{\theta}{2}-1\\ \cos \theta +1&= 2\cos^2 \frac{\theta}{2}  \\ or, 2 \cos^2 \frac{\theta}{2} &=1+\cos \theta \\ or, \cos^2\frac{\theta}{2}  & = \frac{1+\cos \theta}{2}\\ \therefore \ \cos \frac{\theta}{2}  &= \pm\sqrt{\frac{1+\cos \theta}{2}}\\ \end{align}  Second Method \begin{align} \text{ RHS } & = \pm\sqrt{\frac{1+\cos \theta}{2}}\\ \ & =  \pm\sqrt{\frac{1+\left(2\cos^2 \frac{\theta}{2} -1\right)}{2}}\\ \ & =  \pm\sqrt{\frac{1+2\cos^2 \frac{\theta}{2} -1}{2}}\\ \ & =  \pm\sqrt{\frac{2\cos^2 \frac{\theta}{2} }{2}}\\ \ & =  \pm\sqrt{\cos^2 A}\\ \ & = \pm \cos \frac{\theta}{2}  \\ \ & = \cos \frac{\theta}{2} \\ \ & = \text{ LHS } \\ \end{align}


  6. (b) \( \sin \frac{\theta}{2} = \pm \sqrt{\frac{1 - \cos \theta}{2} } \)
Solution:  First Method \begin{align} \cos \theta &= 1-2\sin^2 \frac{\theta}{2}\\ or, 2 \sin^2 \frac{\theta}{2}&=1-\cos \theta \\ or, \sin^2 \frac{\theta}{2} & = \frac{1-\cos \theta }{2}\\ \therefore \ \sin \frac{\theta}{2}&= \pm\sqrt{\frac{1 - \cos \theta }{2}} \end{align}  Second Method  \begin{align} \text{ RHS } & = \pm\sqrt{\frac{1-\cos \theta }{2}}\\ \ & =  \pm\sqrt{\frac{1-(1-2\sin^2 \frac{\theta}{2})}{2}}\\ \ & =  \pm\sqrt{\frac{1-1+2\sin^2 \frac{\theta}{2}}{2}}\\ \ & =  \pm\sqrt{\frac{2\sin^2 \frac{\theta}{2}}{2}}\\ \ & =  \pm\sqrt{\sin^2 \frac{\theta}{2}}\\ \ & = \pm \sin \frac{\theta}{2}\\ \ & = \sin \frac{\theta}{2}\\ \ & = \text{ LHS }  \end{align}

  6. (c) \( \tan \frac{\theta}{2} = \pm \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} \)
Solution:  \begin{align} \text{ RHS } & = \pm\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}\\ \ & =  \pm\sqrt{\frac{1-\left(1-2\sin^2 \frac{\theta}{2}\right)}{1+\left(2cos^2\frac{\theta}{2}+1\right)}}\\ \ & =  \pm\sqrt{\frac{1-1+2\sin^2 \frac{\theta}{2}}{1+2\cos^2\frac{\theta}{2}-1}}\\ \ & =  \pm\sqrt{\frac{2\sin^2 \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}}}\\ \ & =  \pm\sqrt{\tan^2 \frac{\theta}{2}}\\ \ & = \pm \tan \frac{\theta}{2}\\ \ & = \tan \frac{\theta}{2}\\ \ & = \text{ LHS } \\ \end{align}

  6. (d) \( \cot \frac{\theta}{2} = \pm \sqrt{\frac{1+\cos \theta}{1-\cos \theta}} \)
Solution:  \begin{align} \text{ RHS } & = \pm\sqrt{\frac{1+\cos \theta }{1-\cos \theta }}\\ \ & =  \pm\sqrt{\frac{1+\left(2\cos^2 \frac{\theta}{2} - 1 \right)}{1-\left(1- 2\sin^2 \frac{\theta}{2}\right)}}\\ \ & =  \pm\sqrt{\frac{1+2\cos^2 \frac{\theta}{2} - 1}{1-1+2\sin^2 \frac{\theta}{2} }}\\ \ & =  \pm\sqrt{\frac{2\cos^2 \frac{\theta}{2}}{2\sin^2 \frac{\theta}{2}}}\\ \ & =  \pm\sqrt{\cot^2 \frac{\theta}{2}}\\ \ & = \pm \cot \frac{\theta}{2}\\ \ & = \cot \frac{\theta}{2}\\ \ & = \text{ LHS } \\ \end{align}

Prove that:

 7. (a) \( \frac{1+\cos \theta}{2} = \cos^2 \frac{\theta}{2} \)
Solution: \begin{align} \text{LHS } & =\frac{1+\cos \theta }{2} \\ & = \frac{1+2\cos^2 \dfrac{\theta}{2} - 1 }{2}\\ & = \frac{2\cos^2 \dfrac{\theta}{2}  }{2}\\ & = \cos^2 \frac{\theta}{2}\\ & = \text{ RHS} \end{align}

7. (b) \( \frac{1-\cos \theta}{2} = \sin^2 \frac{\theta}{2} \)
Solution:  \begin{align} \text{LHS } & =\frac{1-\cos \theta }{2} \\ & = \frac{1-( 1 - 2\sin^2 \dfrac{\theta}{2} ) }{2}\\ & = \frac{1- 1 + 2\sin^2 \dfrac{\theta}{2} ) }{2}\\ & = \frac{2\sin^2 \dfrac{\theta}{2}  }{2}\\ & = \sin^2 \frac{\theta}{2} \\ & = \text{ RHS} \end{align}

7. (c) \(  \frac{1+\cos \theta }{\sin \theta } = \cot \frac{\theta}{2}  \)
Solution:  \begin{align} \text{LHS } & = \frac{1+\cos \theta }{\sin \theta } \\ & = \frac{1+2\cos^2 \dfrac{\theta}{2} - 1 }{2\sin \dfrac{\theta}{2} \cos \frac{\theta}{2}}\\ & = \frac{2\cos^2\dfrac{\theta}{2}  }{2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} }\\ & = \frac{\cos \dfrac{\theta}{2}   }{\sin \dfrac{\theta}{2} }\\ & = \cot \frac{\theta}{2}  \\ & = \text{ RHS} \end{align}

 7. (d) \(  \frac{1-\cos \theta }{\sin \theta } = \tan \dfrac{\theta}{2} \)
Solution:  \begin{align} \text{LHS } & =\frac{1-\cos \theta }{\sin \theta } \\ & = \frac{1- \left( 1 - 2\sin^2 \dfrac{\theta}{2} \right) }{2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} }\\ & = \frac{1- 1 + 2\sin^2 \dfrac{\theta}{2}  }{2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2}}\\ & = \frac{2\sin^2 \dfrac{\theta}{2}  }{2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} }\\ & = \frac{\sin \dfrac{\theta}{2} }{\cos \dfrac{\theta}{2}}\\ & = \tan \dfrac{\theta}{2} \\ & = \text{ RHS} \end{align} 


7. (e) \(  \frac{1-\cos A }{1+\cos A} = \tan^2 \frac{A}{2} \)
Solution: \begin{align} \text{LHS } & =\frac{1-\cos A }{1+\cos A} \\ & = \frac{1-\left( 1 - 2\sin^2 \dfrac{A}{2} \right) }{1+2\cos^2 \dfrac{A}{2} - 1 }\\ & = \frac{1- 1 + 2\sin^2 \dfrac{A}{2}  }{2\cos^2\dfrac{A}{2} }\\ & = \frac{2\sin^2 \dfrac{A}{2}  }{2\cos^2 \dfrac{A}{2}  }\\ & = \frac{\sin^2 \dfrac{A}{2} }{\cos^2 \dfrac{A}{2}}\\ & = \tan^2 \frac{A}{2} \\ & = \text{ RHS} \end{align}


7. (f) \(  \frac{1+\cos A }{1-\cos A} = \cot^2 \frac{A}{2} \)
Solution:  \begin{align} \text{LHS } & =\frac{1+\cos A }{1-\cos A} \\ & = \frac{1+2\cos^2 \dfrac{A}{2} - 1 }{1-\left( 1 - 2\sin^2 \dfrac{A}{2} \right) }\\ & = \frac{2\cos^2 \dfrac{A}{2}  }{1-1+2\sin^2 \dfrac{A}{2} }\\ & = \frac{2\cos^2 \dfrac{A}{2}  }{2\sin^2 \dfrac{A}{2} }\\ & = \frac{\cos^2 \dfrac{A}{2}}{\sin^2 \dfrac{A}{2}}\\ & = \cot^2 \dfrac{A}{2} \\ & = \text{ RHS} \end{align}

7. (g) \( \frac{\sin \alpha }{1+\cos \alpha } = \tan \dfrac{\alpha}{2} \)
Solution:  \begin{align} \text{LHS } & =\frac{\sin \alpha }{1+\cos 2\alpha} \\ & = \frac{2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} }{1+2\cos^2 \dfrac{\alpha}{2} -1 }\\ & = \frac{2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2}  }{2\cos^2 \dfrac{\alpha}{2} }\\ & = \frac{\sin \dfrac{\alpha}{2} }{\cos \dfrac{\alpha}{2}}\\ & = \tan \dfrac{\alpha}{2} \\ & = \text{ RHS} \end{align}


7. (h) \( \frac{\sin \alpha }{1-\cos \alpha } = \cot \dfrac{\alpha}{2}\)
Solution:  \begin{align} \text{LHS } & =\frac{\sin \alpha }{1-\cos \alpha} \\ & = \frac{2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} }{1-\left( 1 - 2\sin^2 \dfrac{\alpha}{2} \right) }\\ & = \frac{2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2}   }{1-1+2\sin^2 \dfrac{\alpha}{2} }\\ & = \frac{2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2}  }{2\sin^2 \dfrac{\alpha}{2}  }\\ & = \frac{\cos \dfrac{\alpha}{2} }{\sin\dfrac{\alpha}{2}}\\ & = \cot \dfrac{\alpha}{2} \\ & = \text{ RHS} \end{align}


7. (i) \( \frac{1+\sin A}{\cos A} = \frac{\cot \dfrac{A}{2} + 1}{\cot \dfrac{A}{2} - 1 } \)
Solution:  \begin{align} \text{LHS } & = \frac{1+\sin A}{\cos A} \\ & = \frac{\sin^2 \dfrac{A}{2} + \cos^2 \dfrac{A}{2} +2\sin  \dfrac{A}{2}\cos \theta }{\cos ^2\dfrac{A}{2}-\sin^2 \dfrac{A}{2} } \\ & =\frac{\left(\cos\dfrac{A}{2} + \sin \dfrac{A}{2} \right)^2 }{\left(\cos \dfrac{A}{2} - \sin \dfrac{A}{2}\right ) \left(\cos \dfrac{A}{2} + \sin \dfrac{A}{2} \right) }  \\ & = \frac{\cos \dfrac{A}{2} + \sin \dfrac{A}{2} }{ \cos \dfrac{A}{2} - \sin \dfrac{A}{2} } \\ & = \frac{\dfrac{\cos \dfrac{A}{2}}{\sin \dfrac{A}{2}} + \dfrac{\sin \dfrac{A}{2}}{\sin \dfrac{A}{2}} }{ \dfrac{\cos \dfrac{A}{2}}{\sin \dfrac{A}{2}} - \dfrac{\sin \dfrac{A}{2}}{\sin \dfrac{A}{2} } } \\ & =\frac{\cot \dfrac{A}{2} + 1}{\cot \dfrac{A}{2} - 1 }\\ & = \text{ RHS} \end{align}


7. (j) \( \frac{1-\sin A}{\cos A} = \frac{1- \tan \dfrac{A}{2} }{1+\tan \dfrac{A}{2} } \)
Solution:  \begin{align} \text{LHS } & = \frac{1-\sin A}{\cos A} \\ & = \frac{\sin^2 \dfrac{A}{2} + \cos^2 \dfrac{A}{2} -2\sin \dfrac{A}{2} \cos \dfrac{A}{2} }{\cos^2\dfrac{A}{2} -\sin^2 \dfrac{A}{2} } \\ & = \frac{\left(\cos\dfrac{A}{2} - \sin \dfrac{A}{2}\right)^2 }{ \left(\cos\dfrac{A}{2} - \sin \dfrac{A}{2} \right) \left( \cos \dfrac{A}{2} + \sin \dfrac{A}{2} \right) }\\ & = \frac{\cos\dfrac{\dfrac{A}{2}}{2} - \sin \dfrac{\dfrac{A}{2}}{2} }{ \cos \dfrac{\dfrac{A}{2}}{2} + \sin \dfrac{\dfrac{A}{2}}{2} }\\ & = \frac{\dfrac{\cos\dfrac{A}{2}}{\cos \dfrac{A}{2}} - \dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2} } }{ \dfrac{\cos \dfrac{A}{2}}{\cos \dfrac{A}{2}} +\dfrac{ \sin\dfrac{A}{2}}{\cos \dfrac{A}{2}} }\\ & = \frac{1- \tan \dfrac{A}{2} }{1+\tan \dfrac{A}{2} } \\ & = \text{ RHS} \end{align}


7. (k) \( \frac{1-\sin \alpha}{\cos \alpha } = \tan \left(45^\circ - \frac{\alpha}{2} \right) \) 
Solution:  \begin{align} \text{LHS } & = \frac{1-\sin \alpha }{\cos \alpha} \\ & = \frac{\sin^2 \dfrac{\alpha}{2} + \cos^2 \dfrac{\alpha}{2} -2\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} }{\cos^2 \dfrac{\alpha}{2} -\sin^2 \dfrac{\alpha}{2} } \\ & = \frac{\left(\cos \dfrac{\alpha}{2} - \sin \dfrac{\alpha}{2} \right)^2 }{ \left(\cos \dfrac{\alpha}{2} - \sin \dfrac{\alpha}{2} \right) \left( \cos \dfrac{\alpha}{2} + \sin \dfrac{\alpha}{2} \right) }\\ & = \frac{\cos \dfrac{\alpha}{2} - \sin \dfrac{\alpha}{2} }{ \cos \dfrac{\alpha}{2} + \sin \dfrac{\alpha}{2} }\\ & = \frac{\dfrac{\cos \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}} - \dfrac{\sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2} } }{ \dfrac{\cos \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}} +\dfrac{ \sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}} }\\ & = \frac{1- \tan \dfrac{\alpha}{2} }{1+\tan \dfrac{\alpha}{2} } \\ & = \frac{\tan 45^\circ - \tan \dfrac{\alpha}{2} }{1+\tan 45^\circ \tan A }\\ & =\tan \left(45^\circ - \dfrac{\alpha}{2}\right)\\ & = \text{ RHS} \end{align}

7. (l) \( \frac{\cos \alpha }{1-\sin \alpha} = \frac{1+\tan \dfrac{\alpha}{2}}{1-\tan \dfrac{\alpha}{2}} \)
Solution:  \begin{align} \text{LHS } & =  \frac{\cos \alpha }{1-\sin \alpha} \\ & = \frac{\cos^2\dfrac{\alpha}{2} - \sin^2\dfrac{\alpha}{2} }{\cos^2\dfrac{\alpha}{2} + \sin^2\dfrac{\alpha}{2} - 2\sin\dfrac{\alpha}{2} \cos\dfrac{\alpha}{2} }\\ & = \frac{\left( \cos\dfrac{\alpha}{2} - \sin\dfrac{\alpha}{2} \right) \left( \cos\dfrac{\alpha}{2} + \sin\dfrac{\alpha}{2} \right)}{\left( \cos\dfrac{\alpha}{2} - \sin\dfrac{\alpha}{2} \right)^2}\\ & = \frac{\cos\dfrac{\alpha}{2} + \sin\dfrac{\alpha}{2} }{\cos\dfrac{\alpha}{2} - \sin\dfrac{\alpha}{2} }\\ & = \frac{\dfrac{\cos\dfrac{\alpha}{2} }{\cos\dfrac{\alpha}{2} } + \dfrac{\sin\dfrac{\alpha}{2} }{\cos\dfrac{\alpha}{2} }}{\dfrac{\cos\dfrac{\alpha}{2} }{\cos\dfrac{\alpha}{2} } - \dfrac{\sin\dfrac{\alpha}{2} }{\cos\dfrac{\alpha}{2} }}\\ & = \frac{1+\tan\dfrac{\alpha}{2}}{1-\tan\dfrac{\alpha}{2}} \\ & = \text{ RHS} \end{align}


7. (m) \(\frac{\sin^3 \dfrac{\theta}{2} + \cos^3 \dfrac{\theta}{2} }{\sin \dfrac{\theta}{2} + \cos \dfrac{\theta}{2} } = 1 - \frac{1}{2} \sin \theta \)
Solution:  \begin{align} \text{LHS } & =\frac{\sin^3 \dfrac{\theta}{2} + \cos^3 \dfrac{\theta}{2} }{\sin \dfrac{\theta}{2} + \cos \dfrac{\theta}{2} } \\ & =\frac{\left(\sin \dfrac{\theta}{2} + \cos \dfrac{\theta}{2} \right) \left(\sin^2 \dfrac{\theta}{2} - \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} + \cos^2 \dfrac{\theta}{2} \right)}{\sin \dfrac{\theta}{2} + \cos \dfrac{\theta}{2} } \\ & =\sin^2 \dfrac{\theta}{2} - \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} + \cos^2 \dfrac{\theta}{2} \\ & = \sin^2 \dfrac{\theta}{2} + \cos^2 \dfrac{\theta}{2} - \frac{1}{2} (2\sin \dfrac{\theta}{2} \cos\dfrac{\theta}{2} ) \\ & = 1 - \frac{1}{2} \sin \theta  \\ & = \text { RHS }\end{align}

 7. (n) \(\frac{\sin \dfrac{\theta}{2} - \sqrt{1+\sin \theta}}{\cos \dfrac{\theta}{2} -\sqrt{1+\sin \theta}} = \cot \frac{\theta}{2}\)
Solution:  \begin{align} \text{LHS } & = \frac{\sin \dfrac{\theta}{2} - \sqrt{1 +\sin \theta}}{\cos \dfrac{\theta}{2} -\sqrt{1+\sin \theta}}  \\ & = \frac{\sin \dfrac{\theta}{2} - \sqrt{\sin^2\dfrac{\theta}{2} +\cos^2\dfrac{\theta}{2} +2\sin \dfrac{\theta}{2}  \cos \dfrac{\theta}{2} }}{\cos \dfrac{\theta}{2} -\sqrt{\sin^2\dfrac{\theta}{2} +\cos^2\dfrac{\theta}{2} - 2\sin \dfrac{\theta}{2}  \cos \dfrac{\theta}{2} }}  \\ & = \frac{\sin\dfrac{\theta}{2} - \sqrt{\left(\sin\dfrac{\theta}{2}  + \cos\dfrac{\theta}{2} \right)^2 }}{\cos \dfrac{\theta}{2} - \sqrt{\left(\sin\dfrac{\theta}{2}  - \cos\dfrac{\theta}{2} \right)^2 }}\\ & = \frac{\sin\dfrac{\theta}{2}  - \left(\sin\dfrac{\theta}{2}  + \cos\dfrac{\theta}{2}  \right)}{\cos\dfrac{\theta}{2}  - \left(\sin\dfrac{\theta}{2}  + \cos\dfrac{\theta}{2}  \right)}\\ & = \frac{\sin\dfrac{\theta}{2}  - \sin\dfrac{\theta}{2}  - \cos\dfrac{\theta}{2}  }{\cos\dfrac{\theta}{2}  - \sin\dfrac{\theta}{2}  - \cos\dfrac{\theta}{2}  }\\ & = \frac{- \cos \dfrac{\theta}{2} }{- \sin \dfrac{\theta}{2} } \\ & = \frac{\cos \dfrac{\theta}{2} }{ \sin \dfrac{\theta}{2} } \\ & = \cot \frac{\theta}{2} \\ & = \text { RHS }\end{align}


 7. (o) \(\cot 22 \frac{1}{2}^\circ - \tan 22\frac{1}{2}^\circ = 2\)
Solution:  \begin{align} \text{LHS } & =\cot 22 \frac{1}{2}^\circ - \tan 22\frac{1}{2}^\circ  \\ & = \frac{\cos 22 \dfrac{1}{2}^\circ }{\sin 22 \dfrac{1}{2}^\circ } - \frac{\sin 22 \dfrac{1}{2}^\circ }{\cos 22 \dfrac{1}{2}^\circ } \\ & = \frac{\cos^222 \dfrac{1}{2}^\circ -\sin^222 \dfrac{1}{2}^\circ }{\sin 22 \dfrac{1}{2}^\circ \cos22 \dfrac{1}{2}^\circ }\\ & = \frac{\cos 45^\circ}{\sin 22 \dfrac{1}{2}^\circ \cos22 \dfrac{1}{2}^\circ }\\ & = \frac{\cos 45^\circ}{\sin 22 \dfrac{1}{2}^\circ \cos22 \dfrac{1}{2}^\circ }\times \frac{2}{2} \\ & = \frac{2\cos 45^\circ}{2\sin 22 \dfrac{1}{2}^\circ \cos22 \dfrac{1}{2}^\circ }\\ & = \frac{2\cos 45^\circ}{\sin 45^\circ} \\ & = 2\times \frac{\cos 45^\circ}{\sin45^\circ}\\ & = 2\cot45^\circ\\ & = 2\times 1\\ & = 2\\ & = \text { RHS }\end{align}

7. (p) \(\cot 22 \frac{1}{2}^\circ + \tan 22\frac{1}{2}^\circ = 2\sqrt{2} \)
Solution: \begin{align} \text{LHS } & =\cot 22 \frac{1}{2}^\circ + \tan 22\frac{1}{2}^\circ  \\ & = \frac{\cos 22 \dfrac{1}{2}^\circ }{\sin 22 \dfrac{1}{2}^\circ } + \frac{\sin 22 \dfrac{1}{2}^\circ }{\cos 22 \dfrac{1}{2}^\circ } \\ & = \frac{\cos^222 \dfrac{1}{2}^\circ + \sin^222 \dfrac{1}{2}^\circ }{\sin 22 \dfrac{1}{2}^\circ \cos22 \dfrac{1}{2}^\circ }\\ & = \frac{1}{\sin 22 \dfrac{1}{2}^\circ \cos22 \dfrac{1}{2}^\circ }\\ & = \frac{1}{\sin 22 \dfrac{1}{2}^\circ \cos22 \dfrac{1}{2}^\circ }\times \frac{2}{2} \\ & = \frac{2}{2\sin 22 \dfrac{1}{2}^\circ \cos22 \dfrac{1}{2}^\circ }\\ & = \frac{2}{\sin 45^\circ} \\ & = \frac{2}{\dfrac{1}{\sqrt{2}}}\\ & = 2\sqrt{2}\\ & = \text { RHS }\end{align}

Prove that :
8. (a) \( \frac{1+\sin \theta - \cos \theta }{1+\sin \theta + \cos \theta } = \tan \dfrac{\theta}{2}  \)
Solution: \begin{align} \text{LHS } & = \frac{1+\sin \theta - \cos \theta }{1+\sin \theta + \cos \theta }  \\ & = \frac{1+2\sin \dfrac{\theta}{2}   \cos \dfrac{\theta}{2}   - \left(1-2\sin^2 \dfrac{\theta}{2}  \right) }{1+2\sin \dfrac{\theta}{2}   \cos \dfrac{\theta}{2}    + \left(2\cos^2\dfrac{\theta}{2}   -1\right)}  \\ &  = \frac{2\sin \dfrac{\theta}{2}   \cos \dfrac{\theta}{2}   +2\sin^2 \dfrac{\theta}{2}   }{2\sin \dfrac{\theta}{2}   \cos \dfrac{\theta}{2}    + 2\cos^2\dfrac{\theta}{2}   } \\ & = \frac{2\sin \dfrac{\theta}{2}   \left( \cos \dfrac{\theta}{2}   + \sin \dfrac{\theta}{2}   \right) }{2\cos \dfrac{\theta}{2}   \left( \sin \dfrac{\theta}{2}   + \cos \dfrac{\theta}{2}   \right)}\\ & = \frac{\sin \dfrac{\theta}{2}   \left( \sin \dfrac{\theta}{2}   + \cos \dfrac{\theta}{2}   \right) }{\cos \dfrac{\theta}{2}   \left( \sin \dfrac{\theta}{2}   + \cos \dfrac{\theta}{2}   \right)}\\ & = \frac{\sin \dfrac{\theta}{2}   }{\cos \dfrac{\theta}{2}  } \\ & = \tan \dfrac{\theta}{2}   \\ & = \text { RHS }\end{align}


b \( \frac{1+\sin \theta + \cos \theta }{1+\sin \theta  - \cos \theta } = \cot \frac{\theta}{2}  \)
Solution:  \begin{align} \text{LHS } & = \frac{1+\sin \theta + \cos \theta }{1+\sin \theta  - \cos \theta }  \\ & = \frac{1+2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} +\left(2\cos^2 \dfrac{\theta}{2} -1 \right) }{1+2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2}  - \left(1 - 2\sin^2 \dfrac{\theta}{2}\right)}  \\ &  = \frac{2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} +2\cos^2 \dfrac{\theta}{2} }{2\sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2}  + 2\sin^2\dfrac{\theta}{2} } \\ & = \frac{2\cos \dfrac{\theta}{2} \left( \sin \dfrac{\theta}{2} + \cos \dfrac{\theta}{2} \right) }{2\sin \dfrac{\theta}{2} \left( \cos \dfrac{\theta}{2} + \sin \dfrac{\theta}{2} \right)}\\ & = \frac{\cos \dfrac{\theta}{2} \left( \sin \dfrac{\theta}{2} + \cos \dfrac{\theta}{2} \right) }{\sin \dfrac{\theta}{2} \left( \sin \dfrac{\theta}{2} + \cos \dfrac{\theta}{2} \right)}\\ & = \frac{\cos \dfrac{\theta}{2} }{\sin \dfrac{\theta}{2}} \\ & = \cot \dfrac{\theta}{2} \\ & = \text { RHS }\end{align}


8. (c) \( \frac{2\sin A - \sin 2A}{2\sin A + \sin 2A} = \tan^2 \frac{A}{2} \)
Solution:  \begin{align} \text{LHS}& = \frac{2\sin A - \sin 2A}{2\sin A + \sin 2A}  \\ & = \frac{2\sin A - 2\sin A \cos A}{2\sin A + 2\sin A \cos A} \\ & = \frac{2\sin A(1-\cos A)}{2\sin A(1+\cos A)}\\ & = \frac{1-\cos A}{1+\cos A}\\ & = \frac{1-\left( 1 - 2\sin^2 \dfrac{A}{2}\right)}{1+ \left(2\cos^2\dfrac{A}{2} - 1\right)}\\ & = \frac{1- 1 + 2\sin^2 \dfrac{A}{2} }{1+ 2\cos^2\dfrac{A}{2} - 1}\\ & = \frac{ 2\sin^2 \dfrac{A}{2} }{ 2\cos^2\dfrac{A}{2} }\\ & = \frac{\sin^2 \dfrac{A}{2} }{ \cos^2\dfrac{A}{2} }\\ & = \tan^2 \frac{A}{2}\\ & = \text{RHS} \end{align}


8. (d) \( \frac{2\sin A + \sin 2A}{2\sin A - \sin 2A} = \cot^2 \frac{A}{2} \)
Solution: \begin{align} \text{LHS}& = \frac{2\sin A + \sin 2A}{2\sin A - \sin 2A}  \\ & = \frac{2\sin A + 2\sin A \cos A}{2\sin A - 2\sin A \cos A} \\ & = \frac{2\sin A(1+\cos A)}{2\sin A(1-\cos A)}\\ & = \frac{1+\cos A}{1-\cos A}\\ & = \frac{1+ \left(2\cos^2\dfrac{A}{2} - 1\right)}{1-\left( 1 - 2\sin^2 \dfrac{A}{2}\right)}\\ & = \frac{1+ 2\cos^2\dfrac{A}{2} - 1}{1- 1 + 2\sin^2 \dfrac{A}{2} }\\ & = \frac{ 2\cos^2\dfrac{A}{2} }{ 2\sin^2 \dfrac{A}{2} }\\ & = \frac{ \cos^2\dfrac{A}{2} }{ \sin^2 \dfrac{A}{2}}\\ & = \cot^2 \frac{A}{2}\\ & = \text{RHS} \end{align}


8. (e) \( \frac{2\sin A + \sin \dfrac{A}{2}}{1+\cos A + \cos \dfrac{A}{2} } = \tan \frac{A}{2} \)
Solution:  \begin{align} \text{LHS}& = \frac{\sin A + \sin \dfrac{A}{2}}{1+\cos A + \cos \dfrac{A}{2} }  \\ & = \frac{2\sin \dfrac{A}{2} \cos  \dfrac{A}{2} + \sin  \dfrac{A}{2} }{1+2\cos^2 \dfrac{A}{2}  -1 + \cos  \dfrac{A}{2} } \\ & = \frac{2\sin \dfrac{A}{2} \cos  \dfrac{A}{2} + \sin  \dfrac{A}{2} }{2\cos^2 \dfrac{A}{2}  + \cos  \dfrac{A}{2} } \\ & = \frac{\sin  \dfrac{A}{2} \left( 2\cos  \dfrac{A}{2}  + 1\right)} {\cos  \dfrac{A}{2}  \left( 2\cos  \dfrac{A}{2}  + 1\right) }\\ & = \frac{\sin  \dfrac{A}{2} }{\cos  \dfrac{A}{2} }\\ & = \tan  \dfrac{A}{2} \\ & = \text{RHS} \end{align}


8. (f) \( \frac{\sin A}{1+\cos 2A}. \frac{\cos^2 A}{1+\cos A} = \frac{1}{2} \tan \frac{A}{2} \)
Solution: \begin{align} \text{LHS} & = \frac{\sin A}{1+\cos 2A}. \frac{\cos^2 A}{1+\cos A}  \\ & = \frac{\sin A}{1+2\cos^2 A - 1} . \frac{\cos^2 A}{1+2\cos^2  \dfrac{A}{2}  - 1 } \\ & = \frac{\sin A}{2\cos^2 A} . \frac{\cos^2 A}{2\cos^2  \dfrac{A}{2}  } \\ & = \frac{\sin A}{4 \cos^2  \dfrac{A}{2} }\\ & = \frac{2\sin  \dfrac{A}{2}  \cos  \dfrac{A}{2}  }{4\cos^2  \dfrac{A}{2} }\\ & = \frac{1}{2} \times \frac{\sin  \dfrac{A}{2} }{\cos  \dfrac{A}{2} } \\ & = \frac{1}{2} \tan  \dfrac{A}{2}  \\ & = \text{RHS} \end{align}

Prove that:

9. (a) \( \tan \left( \frac{\pi^c}{4} + \frac{A}{2} \right) = \sec A + \tan A \)
Solution:  \begin{align} \text{LHS}& = \tan \left( 45^\circ + \frac{A}{2} \right)  \\ & = \frac{\tan 45^\circ + \tan\dfrac{A}{2}}{1-\tan45^\circ \tan \dfrac{A}{2} }\\ & = \frac{1+\tan \dfrac{A}{2}}{1- 1\times \tan \dfrac{A}{2}} \\ & = \frac{1+\tan \dfrac{A}{2}}{1- \tan \dfrac{A}{2}} \\ & = \frac{1+\dfrac{\sin \dfrac{A}{2} }{\cos \dfrac{A}{2}} }{1-\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}} \\ & = \frac{\dfrac{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} } }{\dfrac{\cos\dfrac{A}{2} - \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} }} \\ & = \frac{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} - \sin\dfrac{A}{2} }\\ & =  \frac{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} - \sin\dfrac{A}{2} }\times  \frac{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }\\ & =  \frac{\left(\cos\dfrac{A}{2} + \sin\dfrac{A}{2}\right)^2 }{\cos^2\dfrac{A}{2} - \sin^2\dfrac{A}{2} }\\ & = \frac{\cos^2\dfrac{A}{2} + 2\cos\dfrac{A}{2} \sin\dfrac{A}{2} + \sin^2\dfrac{A}{2} }{\cos A} \\ & = \frac{\sin^2 \dfrac{A}{2} + \cos^2\dfrac{A}{2} + \sin A }{\cos A } \\ & = \frac{1+\sin A}{\cos A }\\ & = \frac{1}{\cos A} + \frac{\sin A}{\cos A } \\ & = \sec A + \tan A \\ & = \text{RHS } \end{align}

9. (b) \( \tan \left( \frac{\pi^c}{4} + \frac{A}{2} \right) = \sqrt{\frac{1+\sin A}{1-\sin A}} \)
Solution:  \begin{align} \text{LHS}& = \tan \left( 45^\circ + \frac{A}{2} \right)  \\ & = \frac{\tan 45^\circ + \tan\dfrac{A}{2}}{1-\tan45^\circ \tan \dfrac{A}{2} }\\ & = \frac{1+\tan \dfrac{A}{2}}{1- 1\times \tan \dfrac{A}{2}} \\ & = \frac{1+\tan \dfrac{A}{2}}{1- \tan \dfrac{A}{2}} \\ & = \frac{1+\dfrac{\sin \dfrac{A}{2} }{\cos \dfrac{A}{2}} }{1-\dfrac{\sin \dfrac{A}{2}}{\cos \dfrac{A}{2}}} \\ & = \frac{\dfrac{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} } }{\dfrac{\cos\dfrac{A}{2} - \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} }} \\ & = \frac{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} - \sin\dfrac{A}{2} }\\ & =  \frac{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} - \sin\dfrac{A}{2} }\times  \frac{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }{\cos\dfrac{A}{2} + \sin\dfrac{A}{2} }\\ & =  \frac{\left(\cos\dfrac{A}{2} + \sin\dfrac{A}{2}\right)^2 }{\cos^2\dfrac{A}{2} - \sin^2\dfrac{A}{2} }\\ & = \frac{\cos^2\dfrac{A}{2} + 2\cos\dfrac{A}{2} \sin\dfrac{A}{2} + \sin^2\dfrac{A}{2} }{\cos A} \\ & = \frac{\sin^2 \dfrac{A}{2} + \cos^2\dfrac{A}{2} + \sin A }{\cos A } \\ & = \frac{1+\sin A}{\cos A }\\ & = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} \\ & = \sqrt{\frac{(1 + \sin A)^2}{1-\sin^2 A}} \\ & = \sqrt{\frac{(1 + \sin A)^2}{(1-\sin A)(1+\sin A)}} \\ & = \sqrt{\frac{1 + \sin A}{1-\sin A}} \\ & = \text{RHS} \end{align}


9. (c) \( \sec \left( \frac{\pi^c}{4} + \frac{A}{2} \right). \sec \left( \frac{\pi^c}{4} - \frac{A}{2} \right) = 2 \sec A  \)
Solution: \begin{align} \text{LHS}& =\sec \left( \dfrac{\pi^c}{4} + \dfrac{A}{2} \right). \sec \left( \dfrac{\pi^c}{4} - \dfrac{A}{2} \right)  \\ & = \sec \left( 45^\circ + \dfrac{A}{2} \right). \sec \left( 45^\circ - \dfrac{A}{2} \right)  \\ & = \frac{1} { \cos \left( 45^\circ + \dfrac{A}{2} \right). \cos \left( 45^\circ - \dfrac{A}{2} \right) }\\ & = \frac{1} { \cos \left( 45^\circ + \dfrac{A}{2} \right). \cos \left( 45^\circ - \dfrac{A}{2} \right) }  \\ & = \frac{1}{ \left( \cos 45^\circ \cos \dfrac{A}{2} - \sin 45^\circ \sin \dfrac{A}{2} \right) \left( \cos 45^\circ \cos \dfrac{A}{2} + \sin 45^\circ \sin \dfrac{A}{2} \right)}\\ & = \frac{1}{\left(\cos 45^\circ \cos \dfrac{A}{2} \right)^2 - \left(\sin  45^\circ \sin \dfrac{A}{2} \right)^2 }\\ & = \frac{1}{\left(\dfrac{1}{\sqrt{2}} \cos \dfrac{A}{2} \right)^2 - \left(\dfrac{1}{\sqrt{2}} \sin \dfrac{A}{2} \right)^2 }\\ & = \frac{1}{\dfrac{1}{2} \cos^2 \dfrac{A}{2} - \dfrac{1}{2} \sin^2 \dfrac{A}{2}} \\ & = \frac{1}{\dfrac{1}{2} \left( \cos^2 \dfrac{A}{2} - \sin^2 \dfrac{A}{2} \right) } \\ & = \frac{2}{ \cos A } \\ & = 2\sec A \\ & = \text{RHS} \end{align}


9. (d) \( \tan \left( \frac{\pi^c}{4} - \frac{A}{2} \right) = \frac{\cos A}{1+\sin A} \)
Solution:  \begin{align} \text{LHS}& = \tan \left( \frac{\pi^c}{4} - \frac{A}{2} \right) \\ & = \frac{\tan\frac{\pi^c}{4}  - \tan }{1 + \tan\frac{\pi^c}{4} \tan\frac{A}{2} } \\ & = \frac{ 1 - \tan\frac{A}{2} }{1+ 1. \tan \frac{A}{2}}\\ & = \frac{ 1 - \dfrac{\sin \frac{A}{2}}{\cos \frac{A}{2}}}{1+ \dfrac{\sin\frac{A}{2} }{\cos\frac{A}{2} } }\\ & = \frac{\dfrac{\cos \frac{A}{2} - \sin \frac{A}{2} }{\cos \frac{A}{2} }}{ \dfrac{\cos\frac{A}{2} + \sin \frac{A}{2} }{\cos \frac{A}{2} }} \\ & = \frac{ \cos \frac{A}{2} - \sin\frac{A}{2} }{ \cos \frac{A}{2}+ \sin\frac{A}{2} }\\ & = \frac{ \cos \frac{A}{2} - \sin\frac{A}{2} }{ \cos \frac{A}{2}+ \sin \frac{A}{2} }\times  \frac{ \cos \frac{A}{2}+ \sin\frac{A}{2} }{ \cos\frac{A}{2} + \sin\frac{A}{2} } \\ & = \frac{\cos^2\frac{A}{2} - \sin^2\frac{A}{2} }{\left(\cos\frac{A}{2} + \sin\frac{A}{2} \right)^2}\\ & = \frac{\cos \frac{A}{2}}{\sin^2\frac{A}{2} + 2\sin\frac{A}{2} \cos\frac{A}{2} + \cos^2\frac{A}{2} }\\ & = \frac{\cos\frac{A}{2} }{\sin^2 \frac{A}{2}+ \cos^2 \frac{A}{2}+ 2\sin\frac{A}{2} \cos\frac{A}{2} }\\ & = \frac{\cos A }{1+ \sin A }\\ & = \text{RHS} \end{align}


9. (e) \( \cot \left( \frac{A}{2} + 45^\circ \right) - \tan \left( \frac{A}{2} - 45^\circ \right) = \frac{2\cos A}{1+\sin A} \)
Solution: \begin{align} \text{LHS}& = \cot \left( \frac{A}{2} + 45^\circ \right) - \tan \left( \frac{A}{2} - 45^\circ \right) \\ & = \frac{ \cot \cot  - 1}{\cot + \cot } - \frac{\tan - \tan }{1+\tan \tan } \\ & = \frac{\cot \times 1 - 1 }{\cot + 1 } - \frac{\tan  - 1}{1+ 1 \times \tan }\\ & = \frac{\cot - 1}{\cot + 1} - \frac{\tan - 1}{1+\tan }\\ & = \frac{\dfrac{\cos }{\sin } - 1}{\dfrac{\cos }{\sin } + 1 } - \frac{\dfrac{\sin }{\cos } - 1} { 1 + \dfrac{\sin }{\cos }} \\ & = \frac{\dfrac{\cos - \sin }{\sin }} { \dfrac{\cos + \sin }{\sin }}- \frac{ \dfrac{\sin - \cos }{\cos}} { \dfrac{\cos + \sin }{\cos }} \\ & = \frac{\cos - \sin }{\cos + \sin } - \frac{\sin - \cos }{ \cos + \sin } \\ & = \frac{\cos - \sin - \sin + \cos }{\cos + \sin }\\ & = \frac{ 2\cos - 2\sin }{ \cos + \sin }\\ & = \frac{2\left(\cos - \sin \right)}{\cos + \sin } \times \frac{\cos + \sin }{\cos + \sin }\\ & = \frac{2\left( \cos^2 - \sin^2 \right) }{ \left( \cos + \sin \right)^2 }\\ & = \frac{ 2\cos A}{ \cos^2\dfrac{A}{2} + \sin^2 \dfrac{A}{2}+ 2\sin\dfrac{A}{2} \cos \dfrac{A}{2}}\\ & = \frac{ 2\cos A}{1+ \sin A} \\ & = \text{RHS} \end{align}


9. (f) \( \cos^2 \left( \frac{\pi^c}{4} - \frac{A}{4} \right) - \sin^2 \left( \frac{\pi^c}{4} - \frac{A}{4} \right) = \sin \frac{A}{2} \)
Solution: \begin{align} \text{LHS}& = \cos^2 \left( \frac{\pi^c}{4} - \frac{A}{4} \right) - \sin^2 \left( \frac{\pi^c}{4} - \frac{A}{4} \right) \\ & = \cos 2\left( \frac{\pi^c}{4} - \frac{A}{4} \right) \\ & = \cos \left( \frac{\pi^c}{2} - \frac{A}{2}  \right) \\ & = \sin \frac{A}{2}  \\ & = \text{RHS} \end{align}

Prove that:

10. (a) \( \dfrac{ 1 - \tan^2 \left( \dfrac{\pi^c}{4} - \dfrac{\theta}{4} \right)}{1 + \tan^2 \left( \dfrac{\pi^c}{4} - \dfrac{\theta}{4} \right)} = \sin \frac{\theta}{2} \)
Solution:  \begin{align} \text{LHS}& = \dfrac{ 1 - \tan^2 \left( \dfrac{\pi^c}{4} - \dfrac{\theta}{4} \right)}{1 + \tan^2 \left( \dfrac{\pi^c}{4} - \dfrac{\theta}{4} \right)} \\ & = \cos 2\left( \frac{\pi^c}{4} - \frac{\theta}{4} \right)\\ & = \cos \left( \frac{\pi^c}{2} - \frac{\theta}2\right)\\ & = \sin \frac{\theta}{2} \\ & = \text{RHS} \end{align}

10. (b) \( \frac{\sin \theta}{1+\cos 2\theta} \times \frac{2\cos^2 \theta}{1+\cos \theta} = \tan \frac{\theta}{2} \)
Solution:  \begin{align} \text{LHS} & = \frac{\sin \theta}{1+\cos 2\theta}. \frac{2\cos^2 \theta}{1+\cos \theta}  \\ & = \frac{\sin \theta}{1+2\cos^2 \theta - 1} . \frac{2\cos^2 \theta}{1+2\cos^2  \dfrac{\theta}{2}  - 1 } \\ & = \frac{\sin \theta}{2\cos^2 \theta} . \frac{2\cos^2 \theta}{2\cos^2  \dfrac{\theta}{2}  } \\ & = \frac{\sin \theta}{2 \cos^2  \dfrac{\theta}{2} }\\ & = \frac{2\sin  \dfrac{\theta}{2}  \cos  \dfrac{\theta}{2}  }{2\cos^2  \dfrac{\theta}{2} }\\ & =  \frac{\sin  \dfrac{\theta}{2} }{\cos  \dfrac{\theta}{2} } \\ & =  \tan  \dfrac{\theta}{2}  \\ & = \text{RHS} \end{align}


10. (c) \( \tan \left( 45^\circ + \frac{\theta}{2} \right) = \sec \theta + \tan \theta =\sqrt{\frac{1+\sin \theta}{1-\sin \theta} } \)
Solution:  \begin{align} \text{LHS}& = \tan \left( 45^\circ + \frac{\theta}{2} \right)  \\ & = \frac{\tan 45^\circ + \tan\dfrac{\theta}{2}}{1-\tan45^\circ \tan \dfrac{\theta}{2} }\\ & = \frac{1+\tan \dfrac{\theta}{2}}{1- 1\times \tan \dfrac{\theta}{2}} \\ & = \frac{1+\tan \dfrac{\theta}{2}}{1- \tan \dfrac{\theta}{2}} \\ & = \frac{1+\dfrac{\sin \dfrac{\theta}{2} }{\cos \dfrac{\theta}{2}} }{1-\dfrac{\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}}} \\ & = \frac{\dfrac{\cos\dfrac{\theta}{2} + \sin\dfrac{\theta}{2} }{\cos\dfrac{\theta}{2} } }{\dfrac{\cos\dfrac{\theta}{2} - \sin\dfrac{\theta}{2} }{\cos\dfrac{\theta}{2} }} \\ & = \frac{\cos\dfrac{\theta}{2} + \sin\dfrac{\theta}{2} }{\cos\dfrac{\theta}{2} - \sin\dfrac{\theta}{2} }\\ & =  \frac{\cos\dfrac{\theta}{2} + \sin\dfrac{\theta}{2} }{\cos\dfrac{\theta}{2} - \sin\dfrac{\theta}{2} }\times  \frac{\cos\dfrac{\theta}{2} + \sin\dfrac{\theta}{2} }{\cos\dfrac{\theta}{2} + \sin\dfrac{\theta}{2} }\\ & =  \frac{\left(\cos\dfrac{\theta}{2} + \sin\dfrac{\theta}{2}\right)^2 }{\cos^2\dfrac{\theta}{2} - \sin^2\dfrac{\theta}{2} }\\ & = \frac{\cos^2\dfrac{\theta}{2} + 2\cos\dfrac{\theta}{2} \sin\dfrac{\theta}{2} + \sin^2\dfrac{\theta}{2} }{\cos \theta} \\ & = \frac{\sin^2 \dfrac{\theta}{2} + \cos^2\dfrac{\theta}{2} + \sin \theta }{\cos \theta } \\ & = \frac{1+\sin \theta}{\cos \theta }\\ & = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta } \\ & = \sec \theta + \tan \theta \\ & = \text{Mid Term } \end{align}

\begin{align} \text{Mid Term} & = \sec \theta + \tan \theta \\ & = \frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta } \\ & = \frac{1 + \sin \theta}{\cos \theta} \\ & = \sqrt{\frac{(1 + \sin \theta)^2}{\cos^2 \theta}} \\ & = \sqrt{\frac{(1 + \sin \theta)^2}{1-\sin^2 \theta}} \\ & = \sqrt{\frac{(1 + \sin \theta)^2}{(1-\sin \theta)(1+\sin \theta)}} \\ & = \sqrt{\frac{1 + \sin \theta}{1-\sin \theta}} \\ & = \text{RHS} \end{align}

10. (d) \( \sin^6 \frac{\theta}{2} + \cos^6 \frac{\theta}{2}= 1 - \frac{3}{4} \sin^2 \theta \)
Solution:  \begin{align} \text{ LHS} & = \sin^6 \frac{\theta}{2}  +\cos^6 \frac{\theta}{2} \\ & = \left(\sin^2 \frac{\theta}{2} \right)^3 + \left( \cos^2 \frac{\theta}{2} \right)^3 \\ & = \left(\sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2} \right)\left\{\left(\sin^2 \frac{\theta}{2}\right)^2 - \sin^2 \frac{\theta}{2} \cos^2 \frac{\theta}{2} +\left(\cos^2 \frac{\theta}{2} \right)^2 \right\}\\ & = (1) \left\{ \left(\sin^2 \frac{\theta}{2}\right)^2 +2\sin^2 \frac{\theta}{2} \cos^2 \frac{\theta}{2} +\left(\cos^2 \frac{\theta}{2} \right)^2 -3 \sin^2 \frac{\theta}{2} \cos^2 \frac{\theta}{2} \right\}\\ & = \left(\sin^2 \frac{\theta}{2} + \cos^2 \frac{\theta}{2}\right)^2 - \frac{3}{4} \left(2\sin \frac{\theta}{2} \cos \frac{\theta}{2} \right)^2 \\ & = (1)^2 - \frac{3}{4} \left(\sin \theta \right)^2 \\ & = 1 -\frac{3}{4} \sin^2 \theta \\ & = \text{RHS} \end{align}

Prove that:

 11. (a) \((\sin \theta + \sin \beta )^2 +( \cos \theta + \cos \beta )^2 = 4 \cos^2 \frac{\theta - \beta}{2} \)
Solution: \begin{align} \text{LHS}& = (\sin \theta + \sin \beta )^2 +( \cos \theta + \cos \beta )^2 \\ & = \sin^2 \theta + 2\sin \theta \sin \beta + \sin^2 \beta + \cos^2 \theta + 2\cos \theta \cos \beta + \cos^2 \beta \\ & = \sin^2 \theta + \cos^2 \theta + \sin^2 \beta + \cos^2 \beta + 2[\cos \theta \cos \beta + \sin \theta \sin \beta ]\\ & = 1 + 1 + 2\cos (\theta - \beta) \\ & = 2 + 2\left(  2\cos^2 \frac{\theta - \beta}{2} - 1 \right)\\ & = 2 + 4\cos^2 \frac{\theta -\beta}{2} - 2 \\ & =  4\cos^2 \frac{\theta -\beta}{2} \\ & = \text{RHS} \end{align}


 11. (b) \((\cos \theta - \cos \beta )^2 +( \sin \theta - \sin \beta )^2 = 4 \sin^2 \frac{\theta - \beta}{2} \) 

Solution: 
\begin{align} \text{LHS}& = (\cos \theta - \cos \beta )^2 +( \sin \theta - \sin \beta )^2\\ & = \cos^2 \theta - 2\cos \theta \cos \beta + \cos^2 \beta + \sin^2 \theta - 2\sin \theta \sin \beta + \sin^2 \beta  \\ & =\sin^2 \theta + \cos^2 \theta + \sin^2 \beta + \cos^2 \beta - 2\cos \theta \cos \beta - 2\sin \theta \sin \beta \\ & =1+1 - 2(\cos \theta \cos \beta + \sin \theta \sin \beta) \\ & = 2 - 2\cos (\theta + \beta) \\ & = 2 - 2 \left(1-2\sin^2 \frac{\theta - \beta}{2}\right) \\ & = 2 - 2 + 4\sin^2 \frac{\theta - \beta }{2} \\ & =  4\sin^2 \frac{\theta - \beta }{2} \\ & = \text{RHS} \end{align}




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