Multiple Angles

1. (a) Define multiple angle with an example.

Solution:
If A be any angle then 2A, 3A,.... etc are called multiple angles of A.

 

1. (b) Write \( \cos2A \) in terms of \( \cos A \) and \( \sin A \)

Solution:

\[ \cos2A = 2\cos^2 A-1 \]

\[ \cos2A = 1- 2\sin^2 A \]

 

2. (a) Write \( \sin 2A \) in terms of \( \tan A \)

Solution:

\[ \sin 2A = \frac{2 \tan A}{ 1-\tan^2 A } \]

 

2(b) Write \( \tan^2 A \) in terms of \( \cos2A \)

Solution:

\[ \tan^2 A = \frac{1-\cos 2 A }{1+ \cos 2 A } \]

2(c) Write \( \sin3A \) in terms of \( \sin A \).

Solution:

\[ \sin 3A = 3\sin A - 4\sin^3 A \]

2(d) Write \( \tan3θ \) in terms of \( \tan \theta \)

Solution:

\[ \tan 3\theta = \frac{ 3 \tan \theta -\tan^3 \theta}{1-3\tan^2 \theta} \]

3.(a) If \( \sin A = \frac{3}{5} \), find the value of \( \cos 2A \).

Solution:

Given,

\[ \sin A =\frac{3}{5} \]

Now,

\begin{align} \cos 2A &=1-2\sin^2 A\\ \ &=1-2\times\left(\frac{3}{5} \right)^2\\ \ &= 1-\frac{18}{25}\\ \ &=\frac{25-18}{25}\\ \ &=\frac{7}{25}\\ \therefore \cos 2A &= \frac{7}{25} \end{align}

 

3. (b) If \( \sin 2A = \frac{ 24}{25} \) and \( \sin A = \frac{4}{5} \) , then find the value of \( \cos A \).

Solution:

Given,

\[ \sin 2A = \frac{ 24}{25} \]

and

\[ \sin A = \frac{4}{5} \]

We know that,

\begin{align} \sin 2A &=2\sin A \cos A \\ or,\frac{24}{25}&=2\times \frac{4}{5}\times \cos A\\ or,\frac{24}{25}&=\frac{8}{5}\times \cos A\\\\ or, \frac{3}{5}&=\cos A\\ \therefore \cos A &=\frac{3}{5} \\ \end{align}

3(c) If \( \sin A = \frac{4}{5} \), find the value of \( \sin 2A \).

Solution:

Given

\[ \sin A = \frac{4}{5} \]

Now,

\begin{align} \sin 2A &=2\sin A \cos A\\ \ &=2\times \frac{4}{5}\times \frac{3}{5}\\ \ &= \frac{24}{25}\\ \therefore \sin 2A &= \frac{24}{25} \end{align}

3(d) If \( \cos \theta = \frac{12}{13} \), find the value of \( \sin 2\theta \) and \( \cos 2\theta \).

Solution:

Given,

\[ \cos \theta = \frac{12}{13} \]

Now,

\begin{align} \cos 2\theta &=2\cos^2 \theta -1 \\ \ &=2\times\left(\frac{12}{13} \right)^2-1\\ \ &=2\times\left(\frac{144}{169} \right)-1\\ \ &= \frac{288}{169}-1\\ \ &=\frac{288-169}{169}\\ \ &=\frac{119}{169}\\ \therefore \cos 2\theta &= \frac{119}{169} \end{align}

Again,

\begin{align} \sin 2\theta &= \sqrt{1- \cos^2 2\theta}\\ \ &=\sqrt{1-\left(\frac{119}{169}\right)^2}\\ \ &=\sqrt{1-\frac{14161}{28561}}\\ \ &=\sqrt{\frac{28561-14161}{28561}}\\ \ &=\sqrt{\frac{14400}{28561}}\\ \ &=\frac{120}{169}\\ \therefore \sin 2\theta &=\frac{120}{169} \end{align}

 

3(e) If \( \tan \theta = \frac{3}{4} \), find the value of \( \sin 2\theta \) and \( \tan 2\theta \)

Solution:

\begin{align} \sin 2\theta &= \frac{2\tan \theta}{1+\tan^2 \theta }\\ \ &= \frac{2\left( \frac{3}{4} \right)}{1+ \left( \frac{3}{4} \right)^2} \\ \ &=\frac{\frac{6}{4}}{1+\frac{9}{16}}\\ \ & = \frac{\frac{6}{4}}{\frac{16+9}{16}}\\ \ & = \frac{\frac{6}{4}}{\frac{25}{16}}\\ \ & = \frac{6}{4}\times \frac{16}{25}\\ \ & = \frac{24}{25}\\ \therefore \sin 2\theta &= \frac{24}{25}\\ \end{align}

And

\begin{align} \tan 2\theta &= \frac{2\tan \theta}{1-\tan^2 \theta }\\ \ &= \frac{2\left( \frac{3}{4} \right)}{1- \left( \frac{3}{4} \right)^2} \\ \ &=\frac{\frac{3}{2}}{1-\frac{9}{16}}\\ \ & = \frac{\frac{3}{2}}{\frac{16-9}{16}}\\ \ & = \frac{\frac{3}{2}}{\frac{7}{16}}\\ \ & = \frac{3}{2}\times \frac{16}{7}\\ \ & = \frac{24}{7}\\ \therefore \tan 2\theta & = \frac{24}{7}\\ \end{align}

3. (f) If \( \sin \alpha = \frac{1}{2} \), find the value of \( \sin 3\alpha \) and \( \cos 3\alpha \)

Solution:

\begin{align}\sin 3\alpha &=3\sin \alpha -4\sin^3 \alpha \\ \ & = 3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^3\\ \ & = \frac{3}{2}-4\times \frac{1}{8} \\ \ & = \frac{3}{2} - \frac{1}{2}\\ \ & = \frac{3-1}{2} \\ \ & = \frac{2}{2} \\ \ & = 1 \\ \therefore \sin 3\alpha &= 1 \\ \end{align} 

And

\begin{align} \cos 3\alpha &= \sqrt{1- \sin^2 3\alpha}\\ \ &=\sqrt{1-1}\\ \ & = \sqrt{0}\\ \ & = 0\\ \therefore cos 3\alpha &= 0 \\ \end{align}

Alternative

\begin{align} \sin \alpha &= \frac{1}{2} \\ or, \sin \alpha & = \sin 30^{\circ} \\ \therefore \alpha & = 30^{\circ}\ \end{align}

Now

\begin{align} \sin 3\alpha &= \sin 3(30^{\circ})\\ \ &= \sin 90^{\circ} \\ \ & = 1\\ \therefore \sin 3\alpha &=1 \end{align}

And

\begin{align} \cos 3\alpha &= \cos 3(30^{\circ})\\ \ &= \cos 90^{\circ}\\ \ & = 0\\ \therefore \cos 3\alpha &=0 \end{align}

3. (g) If \( \cos \alpha = \frac{ \sqrt{3}}{2} \), find the value of \( \sin 3\alpha \) and \( \cos 3\alpha \)

Solution:

\begin{align} \cos \alpha &= \frac{\sqrt{3}}{2} \\ or, \cos \alpha & = \cos 30^{\circ} \\ \therefore \alpha & = 30^{\circ}\\ \end{align}

Now,

\begin{align} \sin 3\alpha &= \sin 3(30^{\circ})\\ \ &= \sin 90^{\circ} \\ \ & = 1\\ \therefore \sin 3\alpha &=1 \end{align}

And

\begin{align} \cos 3\alpha &= \cos 3(30^{\circ})\\ \ &= \cos 90^{\circ} \\ \ & = 0\\ \therefore \cos 3\alpha &=0 \end{align}

 

3. (h) If \( \tan \beta = \frac{1}{2} \), find the value of \( \tan 3\beta \).

Solution:

Given,

\( \tan \beta = \frac{1}{2} \)

Now,

\begin{align} \tan 3\beta &=\frac{3 \tan \beta -\tan^3 \beta}{1-3\tan^2 \beta}\\ \ &= \frac{3 \left(\frac{1}{2} \right)-\left(\frac{1}{2} \right)^3}{1-3\left(\frac{1}{2} \right)^2}\\ \ & = \frac{\frac{3}{2}-\frac{1}{8}}{1-\frac{3}{4}} \\ \ & = \frac{\frac{12-1}{8}}{\frac{4-3}{4}} \\ \ & = \frac{11}{8} \times \frac{4}{1} \\ \ & = \frac{11}{2}\\ \therefore \tan 3\beta &= \frac{11}{2}\\ \end{align}

4. (a) If \( \cos 2A = \frac{7}{25} \), then show that \( \sin A = \frac{3}{5} \)

Solution:Given,

\( \cos 2A = \frac{7}{25} \)

We know that,

\begin{align} \sin^2 A &=\frac{1-\cos 2A}{2}\\ \ & = \frac{1-\frac{7}{25}}{2}\\ \ & = \frac{\frac{25-7}{25}}{2}\\ \ & = \frac{\frac{18}{25}}{2}\\ \ & = \frac{18}{25}\times \frac{1}{2}\\ \ & = \frac{9}{25} \\ or, \sin^2 A & = \frac{9}{25}\\ or, \sin A & = \sqrt{\frac{9}{25}}\\ \therefore \sin A & = \frac{3}{5} \end{align}

4.(b) If \( \cos 2A = - \frac{ 1}{2} \), then show that \( \cos A = \frac{1}{2} \).

Solution:

Given,

\( \cos 2A = - \frac{ 1}{2} \)

We know that,

\begin{align} \cos^2 A &=\frac{1+\cos 2A}{2}\\ \ & = \frac{1+\left( \frac{-1}{2}\right) }{2}\\ \ & = \frac{\frac{2-1}{2}}{2}\\ \ & = \frac{\frac{1}{2}}{2}\\ \ & = \frac{1}{4}\\ or, \cos^2 A & = \frac{1}{4}\\ or, \cos A & = \sqrt{\frac{1}{4}}\\ \therefore \cos A & = \frac{1}{2}\\ \end{align}

5. (a) Prove that: \( \sin A = \pm \sqrt{\frac{1-\cos 2A}{2}} \)

Solution:

First Method:

\begin{align} \cos 2A &= 1-2\sin^2 A\\ or, 2 \sin^2 A&=1-\cos 2A\\ or, \sin^2 A & = \frac{1-\cos 2A}{2}\\ \therefore \ \sin A&= \pm\sqrt{\frac{1 \cos 2A}{2}}\\ \end{align}

Second Method:

\begin{align} \text{ RHS } & = \pm\sqrt{\frac{1-\cos 2A}{2}}\\ \ & = \pm\sqrt{\frac{1-(1-2\sin^2 A)}{2}}\\ \ & = \pm\sqrt{\frac{1-1+2\sin^2 A}{2}}\\ \ & = \pm\sqrt{\frac{2\sin^2 A}{2}}\\ \ & = \pm\sqrt{\sin^2 A}\\ \ & = \pm \sin A\\ \ & = \sin A\\ \ & = \text{ LHS } \\ \end{align}

5. (b) Prove that: \( \cos A = \pm \sqrt{\frac{1+\cos 2A}{2}} \)

Solution:

First Method:

\begin{align} \cos 2A &= 2\cos^2 A -1\\ \cos 2A +1&= 2\cos^2 A \\ or, 2 \cos^2 A&=1+\cos 2A\\ or, \cos^2 A & = \frac{1+\cos 2A}{2}\\ \therefore \ \cos A&= \pm\sqrt{\frac{1+\cos 2A}{2}}\\ \end{align} 

Second Method:

\begin{align} \text{ RHS } & = \pm\sqrt{\frac{1+\cos 2A}{2}}\\ \ & = \pm\sqrt{\frac{1+(2\cos^2 A-1)}{2}}\\ \ & = \pm\sqrt{\frac{1+2\cos^2 A-1}{2}}\\ \ & = \pm\sqrt{\frac{2\cos^2 A}{2}}\\ \ & = \pm\sqrt{\cos^2 A}\\ \ & = \pm \cos A\\ \ & = \cos A\\ \ & = \text{ LHS } \\ \end{align}

5. (c) Prove that: \( \tan A = \pm \sqrt{\frac{1-\cos 2A}{1+\cos 2A }} \)

Solution:

 \begin{align} \text{ RHS } & = \pm\sqrt{\frac{1-\cos 2A}{1+\cos 2A}}\\ \ & = \pm\sqrt{\frac{1-(1-2\sin^2 A)}{1+(2cos^2A+1)}}\\ \ & = \pm\sqrt{\frac{1-1+2\sin^2 A}{1+2\cos^2A-1}}\\ \ & = \pm\sqrt{\frac{2\sin^2 A}{2\cos^2 A}}\\ \ & = \pm\sqrt{\tan^2 A}\\ \ & = \pm \tan A\\ \ & = \tan A\\ \ & = \text{ LHS } \\ \end{align}

5. (d) Prove that: \( \sec 2A = \frac{\cot^2 A + 1}{ \cot^2 A -1 } \)

Solution:

\begin{align} \text{LHS} & = \sec 2A \\ \ & =\frac{1}{\cos 2A} \\ \ & = \frac{\cos^2 A+ \sin^2 A }{\cos^2 A -\sin^2 A}\\ [\because &\text{ Dividing both sides by } \cos^2 A]\\ \ & =\frac{\frac{\cos^2 A }{\cos^2 A}+\frac{\sin^2 A}{\cos^2 A}}{\frac{\cos^2 A }{\cos^2 A}-\frac{\sin^2 A}{\cos^2 A}}\\ \ & = \frac{\cot^2 A+1}{\cot^2 A -1}\\ \ & = \text{RHS} \end{align}

 

6. (a) \( \frac{\sin 2A}{1+\cos 2A } = \tan A \)

Solution:

\begin{align} \text{ LHS } & = \frac{\sin 2A}{1+\cos 2A }\\ \ & = \frac{2\sin A \cos A }{1+ 2\cos^2 A -1 }\\ \ & = \frac{2 \sin A \cos A }{2\cos^2 A } \\ \ & = \frac{ \sin A }{ \cos A } \\ \ & = \tan A \\ \ & = \text { RHS } \end{align}

6(b) \( \frac{ 1 -\cos 2A }{ \sin 2A } = \tan A \)

Solution:

\begin{align} \text { LHS } & = \frac{ 1 -\cos 2A }{ \sin 2A } \\ \ & = \frac{ 1 - (1 -2\sin^2 A ) } {\sin 2A } \\ \ & = \frac{ 1-1+2\sin^2 A}{2\sin A \cos A } \\ \ & = \frac{ 2\sin^2 A}{2\sin A \cos A} \\ \ & = \frac{ \sin A } { \cos A } \\ \ & = \tan A \\ \ & = \text { RHS } \\ \end{align}

6. (c) \( \frac{\sin 2A}{ 1-\cos 2A} = \cot A \)

Solution:

\begin{align} \text{ LHS } & = \frac{\sin 2A}{ 1-\cos 2A}\\ \ & = \frac{2\sin A \cos A } {1-(1-2\sin^2 A)} \\ \ & = \frac{ 2\sin A \cos A}{1-1+2\sin^2 A}\\ \ & = \frac{ 2\sin A \cos A } {2\sin^2 A } \\ \ & =\frac{ \cos A}{\sin A } \\ \ & = \cot A \\ \ & = \text{ RHS } \\ \end{align}

6. (d) \( \frac{ 1-\cos 2\theta }{1+\cos 2\theta} = \tan^2 \theta \)

Solution:

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

6. (e) \( \frac{1-\tan \alpha}{1+\tan \alpha}=\frac{1-\sin 2\alpha}{\cos 2\alpha} \)

Solution:

 \begin{align} \text{ LHS } & = \displaystyle \frac{1-\tan \alpha}{1+\tan \alpha}\\ \ & = \displaystyle \frac{ 1 - \frac{ \sin \alpha}{\cos \alpha}}{ 1+ \frac{\sin \alpha}{\cos \alpha}}\\ \ & = \frac{\frac{\cos \alpha - \sin \alpha}{\cos \alpha}}{\frac{\cos \alpha + \sin \alpha }{\cos \alpha}} \\ \ & = \frac{ \cos \alpha - \sin \alpha} { \cos \alpha + \sin \alpha } \\ \ & = \frac{ \cos \alpha - \sin \alpha} { \cos \alpha + \sin \alpha } \times \frac{ \cos \alpha - \sin \alpha} { \cos \alpha - \sin \alpha } \\ \ & = \frac{ (\cos \alpha - \sin \alpha)^2 } { \cos^2 \alpha - \sin^2 \alpha } \\ \ & = \frac{\cos^2 \alpha - 2\cos \alpha \sin \alpha + \sin^2 \alpha }{\cos 2\alpha } \\ \ & = \frac{\sin^2 \alpha + \cos^2 \alpha - 2\sin \alpha \cos \alpha }{\cos 2\alpha} \\ \ & = \frac{ 1 - \sin 2\alpha }{\cos 2\alpha }\\ \ & = \text{ RHS } \end{align}

6. (f) \( \frac{ \cos 2\theta}{1+\sin 2\theta } = \frac{ 1-\tan \theta}{1+ \tan \theta} \)

Solution:

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

6. (g) \( \frac{\sin \theta + \sin 2\theta}{1+\cos \theta + \cos 2\theta} = \tan \theta \)

Solution:

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

6. (h) \( \frac{1+ \cos \beta + \cos 2\beta} {\sin \beta + \sin 2\beta } = \cot \beta \)

Solution:

\begin{align} \text{ LHS } & = \frac{ 1+ \cos \beta + \cos 2\beta }{\sin \beta + \sin 2\beta}\\ \ & =\frac{1+\cos \beta + 2\cos^2 \beta-1}{\sin \beta+2\sin \beta \cos \beta }\\ \ & = \frac{\cos \beta( 1+ 2\cos \beta)}{\sin \beta(1+2\cos \beta)}\\ \ & = \frac{\cos \beta }{\sin \beta }\\ \ & = \cot \beta \\ \ & = \text{ RHS } \end{align}

6. (i) \( \frac{1-\sin 2\alpha}{\cos 2\alpha} = \frac{ 1-\tan \alpha}{1+\tan \alpha} = \tan (45^{\circ} - \alpha ) \)

Solution:

\begin{align} \text{ LHS } & = \frac{1-\sin 2\alpha}{\cos 2\alpha} \\ \ & = \frac{\sin^2 \alpha + \cos^2 \alpha -2\sin \alpha \cos \alpha}{\cos^2\alpha -\sin^2 \alpha}\\ \ & =\frac{(\cos \alpha - \sin \alpha )^2}{(\cos \alpha -\sin \alpha)(\cos \alpha + \sin \alpha)} \\ \ & = \frac{\cos \alpha - \sin \alpha }{\cos \alpha + \sin \alpha }\\ \ & = \frac{\frac{\cos \alpha }{\cos \alpha}- \frac{\sin \alpha }{\cos \alpha}}{\frac{\cos \alpha}{\cos \alpha} + \frac{\sin \alpha }{\cos \alpha}}\\ \ & = \frac{1-\tan \alpha }{1+ \tan \alpha } = \text{ Mid Term } \\ \ & = \frac{ \tan 45^{\circ} - \tan \alpha } { 1+ \tan 45^{\circ} \tan \alpha } \\ \ & = \tan( 45^{\circ} - \alpha)\\ \ & = \text{ RHS } \end{align}

 

6. (j) \( \tan \theta + \cot \theta = 2\text{cosec} 2\theta \)

Solution:

\begin{align} \text{LHS } & = \tan \theta+ \cot \theta \\ \ & = \frac{\sin \theta }{\cos \theta}+\frac{\cos \theta}{\sin \theta}\\ \ & = \frac{\sin^2 \theta+\cos^2 \theta}{\cos \theta \sin \theta}\\ \ & = \frac{1}{\sin \theta \cos \theta}\times \frac{2}{2} \\ \ & = \frac{2}{2\sin \theta \cos \theta }\\ \ & = \frac{2}{\sin 2\theta} \\ \ & = 2 \text{ cosec } \theta \\ \ & = \text {RHS} \end{align}

6. (k) \( \text{cosec } 2A - \cot 2A = \tan A \)

Solution:

\begin{align} \text{LHS } & = \text{ cosec } 2A-\cot 2A \\ \ & = \frac{1}{\sin 2A}-\frac{\cos 2A}{\sin 2A}\\ \ & =\frac{1- \cos 2A}{\sin 2A} \\ \ & = \frac{1-(1-2\sin^2 A)}{2\sin A \cos A}\\ \ & = \frac{1-1+2\sin^2 A}{2\sin A \cos A}\\ \ & =\frac{2\sin^2 A}{2\sin A \cos A}\\ \ & =\frac{\sin A}{\cos A }\\ \ & = \tan A \\ \ & = \text{ RHS }\\ \end{align}

7. (a) \( \tan(45^{\circ} + \theta) = \sec 2\theta + \tan 2\theta \)

Solution:

\begin{align} \text{LHS } & = \tan (45^{\circ} + \theta) \\ \ & = \frac{\tan 45^{\circ} + \tan \theta}{1-\tan 45^{\circ} \tan \theta}\\ \ & = \frac{1+\tan \theta}{1-1\times \tan \theta }\\ \ & = \frac{1+\tan \theta}{1- \tan \theta }\\ \ & = \frac{1+\frac{\sin \theta}{\cos\theta}}{1- \frac{\sin \theta}{\cos \theta} }\\ \ & = \frac{\frac{\cos \theta + \sin \theta }{\cos \theta}}{\frac{\cos \theta - \sin \theta }{\cos \theta}}\\ \ & = \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta }\\ \ & = \frac{\cos \theta + \sin \theta}{\cos \theta - \sin \theta } \times \frac{\cos \theta + \sin \theta}{\cos \theta + \sin \theta }\\ \ & = \frac{(\cos \theta + \sin \theta )^2 }{\cos^2 \theta - \sin^2 \theta }\\ \ & = \frac{ \cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta } {\cos 2\theta}\\ \ & = \frac{ \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta }{ \cos 2\theta} \\ \ & = \frac{ 1+ \sin 2\theta } { \cos 2\theta} \\ \ & = \frac{ 1}{\cos 2\theta} + \frac{ \sin 2\theta}{\cos 2\theta}\\ \ & = \sec 2\theta + \tan 2\theta \\ \ & = \text{ RHS} \end{align}

7. (b) \( 1 -\sin 2A = 2 \sin^2 (45^{\circ} - A) \)

Solution:

\begin{align} \text{RHS } \ & = 2\sin^2(45^{\circ}-A) \\ \ & =2\times\frac{1-\cos2(45^{\circ}-A)}{2}\\ \ & [\because \sin^2 \theta = \frac {1-\cos 2\theta}{2} ]\\ \ & = 1- \cos ( 90^{\circ} - 2A) \\ \ & = 1-\sin 2A \\ \ & = \text{ LHS }\\ \end{align}

7. (c) \( 2\cos^2 (45^{\circ} -\theta ) = 1+ \sin 2\theta \)

Solution:

\begin{align} \text{ LHS } & = 2 \cos^2 (45^{\circ} -\theta)\\ \ & = 2 \times \frac{1+\cos 2(45^{\circ}-\theta)}{2}\\ \ & \left[ \because \cos^2 A = \frac{ 1+\cos 2A}{2} \right]\\ \ & = 1+\cos(90^{\circ} -2\theta)\\ \ & = 1+\sin 2\theta\\ \ & = \text{ RHS} \end{align}

7. (d) \( \cos^2 ( 45^{\circ} -A) -\sin^2 (45^{\circ} -A) = \sin 2A \)

Solution:

\begin{align}\text{LHS } & = \cos^2 ( 45^{\circ} -A) -\sin^2 (45^{\circ} -A)\\ \ & = \cos 2(45^{\circ} - A) \\ \ & = \cos (90^{\circ} - 2A) \\ \ & = \sin 2A \\ \ & = \text { RHS }\end{align}

7. (e) \( \tan (A + 45^{\circ}) - \tan ( A - 45^{\circ}) = \frac{2(1+\tan^2 A)}{1-\tan^2 A} \)

Solution:

\begin{align}\text{LHS }\ & = \tan(A+45^{\circ}-\tan (A-45^{\circ}) \\ \ & =\frac{\tan A+\tan45^{\circ}}{1-\tan A\tan45^{\circ}}-\frac{\tan A-\tan45^{\circ}}{1+\tan A\tan45^{\circ}}\\ \ & = \frac{\tan A+1}{1-\tan A \times 1}- \frac{\tan A-1}{1+\tan A \times 1}\\ \ & =\frac{\tan A+1}{1-\tan A}-\frac{\tan A-1}{1+\tan A}\\ \ & =\frac{ (\tan A+1) (1+\tan A) -( \tan A-1)(1-\tan A)}{(1-\tan A)(1+\tan A) } \\ \ & = \frac{\tan A+\tan^2 A+1+\tan A - (\tan A -\tan^2 A -1+\tan A)}{1^2-\tan^2 A} \\ \ & =\frac{\tan A+\tan^2 A+1+\tan A- \tan A + \tan^2 A+1 -\tan A}{1-\tan^2 A}\\ \ & =\frac{2+2\tan^2 A}{1-\tan^2 A} \\ \ & =\frac{2\left(1+\tan^2A\right)}{1-\tan^2A}\\ \ & = \text{ RHS } \\ \end{align}

7. (f) \( \frac{1+\tan^2 (45^{\circ}-\theta )}{1-\tan^2 (45^{\circ}-\theta )} = \text{ cosec } 2\theta \)

Solution:

\begin{align} \text{LHS } \ & = \frac{1+\tan^2 (45^{\circ}-\theta )}{1-\tan^2 (45^{\circ}-\theta )}\\ \ & = { \displaystyle \frac{1}{\frac{1-\tan^2(45^{\circ}-\theta)}{1+\tan^2(45^{\circ}-\theta)}} }\\ \ & = {\cos 2(45^{\circ}-\theta ) } \\ \ & = \frac{1}{\cos (90^{\circ} - 2\theta} ) \\ \ & = \frac{1}{\sin 2\theta } \\ \ & = \text{cosec } 2\theta \\ \ & =\text{ RHS } \end{align}

8. (a) If \( \cos\theta=\frac{1}{2}\left(a+\frac{1}{a}\right) \) then prove that: \( \cos2\theta=\frac{1}{2}\left(a^2+\frac{1}{a^2}\right). \)

Solution:

\begin{align} \text{LHS } &=\cos2\theta\\ \ &=2\cos^2\theta-1\\ \ &=2\left(\cos\theta\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}

8. (b) If \( \sin\theta=\frac{1}{2}\left(b+\frac{1}{b}\right) \), prove that:\(  \cos2\theta=-\frac{1}{2}\left(b^2+\frac{1}{b^2}\right) \)

Solution: \begin{align} \text{LHS } \ & \displaystyle =\cos2\theta\\ \ & \displaystyle=1-2\sin^2\theta\\ \ & \displaystyle =1-2\left(\sin\theta\right)^{^2}\\ \ & \displaystyle =1-2\left\{\frac{1}{2}\left(b+\frac{1}{b}\right)\right\}^2\\ \ & \displaystyle =1-2\times\frac{1}{4}\left(b+\frac{1}{b}\right)^2\\  \ & \displaystyle =1-\frac{1}{2}\left(b^2+2\times b\times\frac{1}{b}+\frac{1}{b^2}\right)\\ \ & \displaystyle =1-\frac{1}{2}\left(b^2+2+\frac{1}{b^2}\right)\\ \ & \displaystyle =1-\frac{1}{2}\left(b^2+\frac{1}{b^2}+2\right)\\ \ & \displaystyle =1-\frac{1}{2}\left(b^2+\frac{1}{b^2}\right)-\frac{1}{2}\times2\\ \ & \displaystyle =1-\frac{1}{2}\left(b^2+\frac{1}{b^2}\right)-1\\ \ & \displaystyle =-\frac{1}{2}\left(b^2+\frac{1}{b^2}\right)\\ \ & \displaystyle =\text{ RHS } \end{align}

8. (c) If \( \sin\beta=\frac{1}{2}\left(k+\frac{1}{k}\right) \) , show that \( \sin3\beta=-\frac{1}{2}\left(k^3+\frac{1}{k^3}\right) \)

Solution:

\begin{align} \text{LHS } \ & =\sin3\beta\\ \ & =3\sin\beta-4\sin^3\beta\\  \ & =3\times\frac{1}{2}\left(k+\frac{1}{k}\right)-4\times\left\{\frac{1}{2}\left(k+\frac{1}{k}\right)\right\}^3\\  \ & =\frac{3}{2}\left(k+\frac{1}{k}\right)-4\times\frac{1}{8}\left(k+\frac{1}{k}\right)^3\\ \ & =\frac{3}{2} \left( k + \frac{1}{k} \right) - \frac{1}{2} \left(k+\frac{1}{k}\right)^3\\ \ & =\frac{3}{2} \left( k + \frac{1}{k} \right) - \frac{1}{2} \left\{k^3+ \frac{1}{k^3}+3\times k\times \frac{1}{k} \left( 1+ \frac{1}{k} \right) \right\}\\ \ & =\frac{3}{2} \left( k + \frac{1}{k} \right) - \frac{1}{2} \left\{k^3+ \frac{1}{k^3}+3 \left( k + \frac{1}{k} \right) \right\}\\ \ & =\frac{3}{2} \left( k + \frac{1}{k} \right) - \frac{1}{2} \left(k^3+ \frac{1}{k^3}\right) - \frac{3}{2} \left( k+ \frac{1}{k} \right) \\ \ & = - \frac{1}{2} \left(k^3+ \frac{1}{k^3}\right) \\ \ & =\text{ RHS}\\ \end{align}

Alternative

\begin{align} \text{LHS } \ & =\sin3\beta\\ \ & =3\sin\beta-4\sin^3\beta\\  \ & =3\times\frac{1}{2}\left(k+\frac{1}{k}\right)-4\times\left\{\frac{1}{2}\left(k+\frac{1}{k}\right)\right\}^3\\  \ & =\frac{3}{2}\left(k+\frac{1}{k}\right)-4\times\frac{1}{8}\left(k+\frac{1}{k}\right)^3\\  \ & =\frac{1}{2}\left\{3\left(k+\frac{1}{k}\right)-\left(k+\frac{1}{k}\right)^3\right\}\\   \ & =-\frac{1}{2}\left\{\left(k+\frac{1}{k}\right)^3-3\left(k+\frac{1}{k}\right)\right\}\\ \ & =-\frac{1}{2}\left\{\left(k+\frac{1}{k}\right)^3-3\times k\times\frac{1}{k}\left(a+\frac{1}{k}\right)\right\}\\ \ & =-\frac{1}{2}\left(k^3+\frac{1}{k^3}\right) \left[\because k^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\right]\\ \ & =\text{ RHS}\\ \end{align}

9. (a) \( (\cos 2A -\cos 2B)^2 + (\sin 2A + \sin 2B)^2 = 4\sin^2 (A+B) \)

Solution:

 \begin{align} \text{LHS } & = (\cos 2A -\cos 2B)^2 + (\sin 2A + \sin 2B)^2 \\ \ & = \cos^2 2A -2\cos 2A \cos 2B + \cos ^2 2B + \sin^2 2A +2\sin 2A \sin 2B + \sin^2 2B \\ \ & = \sin^2 2A + \cos^2 2A + \sin^2 2B+ \cos^2 2B -2(\cos 2A \cos 2B - \sin 2A \sin 2B) \\ \ & = 1+1-2\cos (2A+2B) \\ \ & = 2 -2\cos (2A+2B) \\ \ & = 2 -2\cos 2(A+B) \\ \ & = 2 - 2\{ 1-2\sin^2 (A+B) \}\\ \ & = 2 -2+4\sin^2(A+B)\\ \ & = 4\sin^2(A+B)\\ \ & = \text{ RHS } \end{align}

9. (b) \( (\sin 2A - \sin 2B)^2 + (\cos 2A + \cos 2B)^2 = 4\cos^2(A+B) \)

Solution:

\begin{align} \text{LHS } & =(\sin 2A - \sin 2B)^2+(\cos 2A +\cos 2B)^2 \\ \ & = \sin^2 2A -2\sin 2A \sin 2B + \sin^2 2B +\cos^2 2A +2\cos 2A \cos 2B + \cos ^2 2B \\ \ & = \sin^2 2A + \cos^2 2A + \sin^2 2B+ \cos^2 2B + 2(\cos 2A \cos 2B - \sin 2A \sin 2B) \\ \ & = 1+1+2\cos (2A+2B) \\ \ & = 2 +2\cos (2A+2B) \\ \ & = 2 +2\cos 2(A+B) \\ \ & = 2 +2\{ 2\cos^2 (A+B)-1 \}\\ \ & = 2 +4\cos^2(A+B)-2\\ \ & = 4\cos^2(A+B)\\ \ & = \text{ RHS } \end{align}

10. (a) Prove that: \( \cos^2\theta+\sin^2\theta.\cos2\alpha=\cos^2\alpha+\sin^2\alpha.\cos2\theta \)

Solution:

\begin{align} \text{ LHS} \ & =\cos^2\theta+\sin^2\theta.\cos2\alpha\\ \ & =\cos^2\theta+\sin^2\theta\left(1-2\sin^2\alpha\right)\\ \ & =\cos^2\theta+\sin^2\theta-2\sin^2\theta\sin^2\alpha\\ \ & =1-2\sin^2\theta\sin^2\alpha\\ \ & =\cos^2\alpha+\sin^2\alpha-2\sin^2\theta\sin^2\alpha\\ \ & =\cos^2\alpha+\sin^2\alpha\left(1-2\sin^2\theta\right)\\ \ & =\cos^2\alpha+\sin^2\alpha\cos2\theta\\ \ & = \text{ RHS}\\ \end{align}

10. (b) \( (1+\cos 2\theta + \sin 2\theta )^2 =4\cos^2 \theta(1+\sin 2\theta) \)

Solution:

\begin{align} \text{LHS} & = (1+\cos 2\theta + \sin 2\theta )^2 =4\cos^2 \theta(1+\sin 2\theta)\\ \ & = ( 1+ 2\cos^2 \theta -1 + 2\sin \theta \cos \theta )^2\\ \ & = (2\cos^2 \theta + 2\sin \theta \cos \theta)^2 \\ \ & = \{ 2\cos \theta(\cos \theta + \sin \theta ) \}^2\\ \ & = 4\cos^2 \theta ( \cos \theta +\sin \theta)^2 \\ \ & = 4\cos^2 \theta (\cos^2 \theta + 2\cos \theta \sin \theta + \sin^2 \theta )\\ \ & = 4\cos^2 \theta ( \sin^2 \theta + \cos^2 \theta + 2\cos \theta \sin \theta )\\ \ & = 4\cos^2 \theta (1+\sin 2\theta)\\ \ & = \text { RHS} \end{align}

10. (c) Prove that: \( (2\cos A+1)(2\cos A-1)\left(2\cos2A-1\right)=1+ 2\cos4A \)

Solution:

\begin{align} \text{LHS} \ & =(2\cos A+1)(2\cos A-1)\left(2\cos2 A-1\right)\\ \ & =\left\{\left(2\cos A\right)^2-1^2\right\}\left(2\cos2 A-1\right)\\ \ & =\left(4\cos^2 A-1\right)\left(2\cos2 A-1\right)\\ \ & =\left(4\times\frac{1+\cos2 A}{2}-1\right)\left(2\cos2 A-1\right)\\ \ & =\left(2+2\cos2 A-1\right)\left(2\cos2 A-1\right)\\ \ & =\left(2\cos2 A+1\right)\left(2\cos2 A-1\right)\\ \ & =\left(2\cos2 A\right)^2-1^2\\ \ & =4\cos^22 A-1\\ \ & =4\times\frac{1+\cos4 A}{2}-1\\ \ & =2\left(1+\cos4 A\right)-1\\ \ & =2+2\cos4 A-1\\ \ & =1+ 2\cos4 A\\ \ & = \text{ RHS }\\ \end{align}

10. (d) \( 1+\cos 8 \theta = ( 2\cos 4\theta -1 ) (2\cos 2\theta -1)( 2\cos \theta -1) (2\cos \theta +1) \)

Solution:

\( \begin{align} \text{RHS} \ & = (2\cos 4\theta-1)(2\cos 2\theta -1) ( 2\cos 2\theta -1)(2\cos \theta +1)\\ \ & = (2\cos 4\theta-1)(2\cos 2\theta -1)\left\{(2\cos 2\theta )^2 -1^2 \right\}\\ \ & = (2\cos 4\theta-1)(2\cos 2\theta -1)(4\cos^2\theta -1^2 )\\ \ & = (2\cos 4\theta-1)(2\cos 2\theta -1) \left( 4\times \frac{1+\cos 2\theta}{2} -1\right)\\ \ & = (2\cos 4\theta-1)(2\cos 2\theta -1) (2+2\cos 2\theta - 1)\\ \ & = (2\cos 4\theta-1)(2\cos 2\theta -1) (2\cos 2\theta +1 ) \\ \ & = (2\cos 4\theta -1 ) \{ (2\cos 2\theta)^2 -1^2 \}\\ \ & = (2\cos 4\theta -1 ) (4\cos^2 2\theta-1)\\ \ & = (2\cos 4\theta -1 )\left( 4\times \frac{1+\cos 4\theta}{2}-1 \right) \\ \ & = (2\cos 4\theta -1 ) ( 2+ 2\cos 4\theta -1 ) \\ \ & = ( 2\cos 4\theta -1 ) ( 2\cos 4\theta +1)\\ \ & = (2\cos 4\theta )^2 -1^2\\ \ & = 4\cos^2 4 \theta -1\\ \ & = 4 \times \frac{1+\cos 8 \theta}{2} -1\\ \ & = 2+2\cos 8 \theta -1\\ \ & = 1+\cos 8 \theta\\ \ & = \text{ LHS }\\ \end{align} \)

11. (a) Prove that: \( \cos^6\theta-\sin^6\theta=\frac{1}{4}\left(\cos^32\theta+3\cos2\theta\right) \)

Solution:

\begin{align} \text{LHS } \ & =\cos^6\theta-\sin^6\theta\\ \ & =\left(\cos^2\theta\right)^3-\left(\sin^2\theta\right)^3\\ \ & =\left(\cos^2\theta-\sin^2\theta\right)\left\{\left(\cos^2\theta\right)^2+\cos^2\theta\sin^2\theta+\left(\sin^2\right)^2\right\}\\\ & =\cos2\theta\left\{\left(\cos^2\theta\right)^2+2\cos^2\theta\sin^2\theta+\left(\sin^2\right)^2-\cos^2\theta\sin^2\theta\right\}\\ \ & =\cos2\theta\left\{\left(\cos^2\theta+\sin^2\theta\right)^2-\frac{1}{4}\left(2\sin\theta\cos\theta\right)^2\right\}\\ \ & =\cos2\theta\left\{\left(1\right)^2-\frac{1}{4}\left(\sin2\theta\right)^2\right\}\\ \ & =\cos2\theta\left\{1-\frac{1}{4}\sin^22\theta\right\}\\ \ & =\cos2\theta\left\{\frac{4-\sin^22\theta}{4}\right\}\\ \ & =\cos2\theta\left\{\frac{4-\left(1-\cos^22\theta\right)}{4}\right\}\\ \ & =\cos2\theta\left(\frac{4-1+\cos^22\theta}{4}\right)\\ \ & =\cos2\theta\left(\frac{3+\cos^22\theta}{4}\right)\\ \ & =\frac{3\cos2\theta+\cos^32\theta}{4}\\ \ & =\frac{1}{4}\left(\cos^32\theta+3\cos2\theta\right)\\ \ & = \text{ RHS }\\ \end{align}

11. (b) Prove that: \( \cos^6\theta+\sin^6\theta=\frac{1}{8}\left(5+3\cos4\theta\right) \)

Solution:

\begin{align} \text{LHS } \ & =\cos^6\theta+\sin^6\theta\\ \ & =\left(\cos^2 \theta \right)^3+\left(\sin^2\theta \right)^3\\ \ & =\left(\sin^2\theta+\cos^2\theta\right)\left\{\left(\cos^2\right)^2-\cos^2\theta\sin^2\theta+\left(\sin^2\theta\right)^2\right\}\\ \ & =\left(1\right)\left\{\left(\cos^2\theta\right)^2+2\cos^2\theta\sin^2\theta+\left(\sin^2\theta\right)^2-3\cos^2\theta\sin^2\theta\right\}\\ \ & =\left\{\left(\cos^2\theta+\sin^2\theta\right)^2-\frac{3}{4}\left(2\cos\theta\sin\theta\right)^2\right\}\\ \ & =\left(1\right)^2-\frac{3}{4}\left(\sin2\theta\right)^2\\ \ & =1-\frac{3}{4}\sin^22\theta\\ \ & =1-\frac{3}{4}\times\frac{1-\cos4\theta}{2}\\ \ & =\frac{8-3+3\cos4\theta}{8}\\ \ & =\frac{5+3\cos4\theta}{8}\\ \ & =\frac{1}{8}\left(5+3\cos4\theta\right)\\ \ & = \text{ RHS}\\ \end{align}

11. (c) Prove that: \( \cos^8\theta+\sin^8\theta=1-\sin^22\theta+\frac{1}{8}\sin^42\theta \)

Solution:

\begin{align} \text{LHS } \ & =\cos^8\theta+\sin^8\theta\\ \ & =\left(\cos^4\theta\right)^2+\left(\sin^4\theta\right)^2\\ \ & =\left(\cos^4\theta-\sin^4\theta\right)^2+2\cos^4\theta\sin^4\theta\\ \ & =\left\{\left(\cos^2\theta\right)^2-\left(\sin^2\theta\right)^2\right\}^2+\frac{2^4}{2^3}\cos^4\theta\sin^4\theta\\ \ & =\left\{\left(\cos^2\theta+\sin^2\theta\right)\left(\cos^2\theta-\sin^2\theta\right)\right\}^2 +\frac{1}{8}\left(2\sin\theta\cos\theta\right)^4\\ \ & =\left\{1\times\cos2\theta\right\}^2+\frac{1}{8}\left(\sin2\theta\right)^4\\ \ & =\cos^22\theta+\frac{1}{8}\sin^42\theta\\ \ & =1-\sin^22\theta+\frac{1}{8}\sin^42\theta\\ \ & = \text{ RHS} \end{align}

11. (d) Prove that: \( \frac{1}{\sin10^\circ}-\frac{\sqrt{3}}{\cos10^\circ} \)

Solution:

\begin{align} \text{LHS } \ & =\frac{1}{\sin10^\circ}-\frac{\sqrt{3}}{\cos10^\circ}\\ \ & =\frac{\cos10^\circ-\sqrt{3}\sin10^\circ}{\sin10^\circ\cos10^\circ}\\ \ & =\frac{\frac{1}{2}\cos10^\circ-\frac{\sqrt{3}}{2}\sin10^\circ}{\frac{1}{2}\sin10^\circ\cos10^\circ}\\ \ & =\frac{\sin30^\circ\cos10^\circ-\cos30^\circ\sin10^\circ}{\frac{1}{2}\sin10^\circ\cos10^\circ}\\ \ & =\frac{\sin\left(30^\circ-10^\circ\right)}{\frac{1}{2}\sin10^\circ\cos10^\circ}\\ \ & =\frac{\sin20^\circ}{\frac{1}{2}\sin10^\circ\cos10^\circ}\\ \ & =\frac{2\sin10^\circ\cos10^\circ}{\frac{1}{2}\sin10^\circ\cos10^\circ}\\ \ & =\frac{2}{\frac{1}{2}}\\ \ & =2\times2\\ \ & =4\\ \ & = \text{ RHS}\\ \end{align}

11. (e) Prove that: \( \sqrt{3}\text{cosec }40^{\circ}+\sec40^{\circ}=4 \)

Solution:

\begin{align} \text{LHS } \ & =\sqrt{3}\text{cosec }40^\circ+\sec40^\circ\\ \ & =\frac{\sqrt{3}}{\sin40^\circ}+\frac{1}{\cos40^\circ}\\ \ & =\frac{\sqrt{3}\cos40^\circ+\sin40^\circ}{\sin40^\circ\cos40^\circ}\\ \ & =\frac{\frac{\sqrt{3}}{2}\cos40^\circ+\frac{1}{2}\sin40^\circ}{\frac{1}{2}\sin40^\circ\cos40^\circ}\\ \ & =\frac{\sin60^\circ\cos40^\circ+\cos60^\circ\sin60^\circ}{\frac{1}{2}\sin40^\circ\cos40^\circ}\\ \ & =\frac{\sin100^\circ}{\frac{1}{2}\sin40^\circ\cos40^\circ}\\ \ & =\frac{\sin\left(180^\circ-80^\circ\right)}{\frac{1}{2}\sin40^\circ\cos40^\circ}\\ \ & =\frac{\sin80^\circ}{\frac{1}{2}\sin40^\circ\cos40^\circ}\\ \ & =\frac{2\sin40^\circ\cos40^\circ}{\frac{1}{2}\sin40^\circ\cos40^\circ}\\ \ & =2\times2\\ \ & =4\\ \ & = \text{ RHS } \end{align}

11. (f) Prove that:\( \sqrt{3}\text{cosec }20^{\circ}-\sec20^{\circ}=4 \)

Solution:

\begin{align} \text{ LHS } = \ & =\sqrt{3}\text{cosec }20^{\circ}-\sec20^{\circ}\\ \ & =\frac{\sqrt{3}}{\sin20^\circ}-\frac{1}{\cos20^\circ}\\ \ & =\frac{\sqrt{3}\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ}\\ \ & =\frac{\frac{\sqrt{3}}{2}\cos20^\circ-\frac{1}{2}\sin20^\circ}{\frac{1}{2}\sin20^\circ\cos20^\circ}\\ \ & =\frac{\sin60^\circ\cos20^\circ-\cos60^\circ\sin20^\circ}{\frac{1}{2}\sin20^\circ\cos20^\circ}\\ \ & =\frac{\sin\left(60^\circ-20^\circ\right)}{\frac{1}{2}\sin20^\circ\cos20^\circ}\\ \ & =\frac{\sin40^\circ}{\frac{1}{2}\sin20^\circ\cos20^\circ}\\ \ & =\frac{2\sin20^\circ\cos20^\circ}{\frac{1}{2}\sin20^\circ\cos20^\circ}\\ \ & =\frac{2}{\frac{1}{2}}\\ \ & =4\\ \ & = \text{ RHS } \end{align}

12. (a) \( \frac{1}{\sin 2A} + \frac{ \cos 4A}{\sin 4A} = \cot A - \text{cosec } 4A \)

Solution:

\begin{align} \text{LHS } & = \frac{1}{\sin 2A} + \frac{ \cos 4A}{\sin 4A} \\ \ & = \frac{1}{\sin 2A} + \frac{ 2\cos^2 2A -1}{2\sin 2A \cos 2A} \\ \ & = \frac{2\cos 2A + 2\cos^2 2A -1 }{2\sin 2A \cos 2A} \\ \ & = \frac{ 2\cos 2A(1+\cos 2A) - 1}{2\sin 2A \cos 2A } \\ \ & = \frac{2\cos 2A (1+\cos 2A}{2\sin 2A \cos 2A} - \frac{1}{2\sin 2A \cos 2 A }\\ \ & = \frac{ 1+\cos 2A}{\sin 2A} - \frac{1}{\sin 4A} \\ \ & = \frac{ 1+ 2\cos^2 A -1}{2\sin A \cos A} - \text{cosec } 4A\\ \ & = \frac{2\cos^2 A }{2\sin A \cos A} -\text{cosec } 4A\\ \ & = \frac{ \cos A } {\sin A} -\text{cosec } 4A\\ \ & = \cot A - \text{cosec } 4A\\ \ & = \text{ RHS } \\ \end{align}

12. (b)\( \cot 8A +\text{cosec } 4A = \cot 2A - \text{cosec }8A \)

Solution:

\begin{align} \text{LHS } & = \cot 8A +\text{cosec } 4A\\ \ & = \frac{\cos 8A}{\sin 8A}+\frac{1}{\sin 4A}\\ \ & = \frac{2\cos^2 4A-1 }{2\sin 4A \cos 4A}+\frac{1}{\sin 4A}\\ \ & = \frac{2\cos^2 4A-1+2\cos 4A}{2\sin 4A \cos 4A}\\ \ & = \frac{2\cos^2 4A+2\cos 4A-1}{2\sin 4A \cos 4A}\\ \ & = \frac{2\cos 4A(\cos 4A+1)-1}{2\sin 4A \cos 4A}\\ \ & = \frac{2\cos 4A(\cos 4A+1)}{2\sin 4A \cos 4A} -\frac{1}{2\sin 4A \cos 4A}\\ \ & = \frac{\cos 4A+1}{\sin 4A } -\frac{1}{\sin 8A}\\ \ & = \frac{2\cos^2 2A-1+1}{2\sin 2A \cos 2A } -\text{cosec }{8A}\\ \ & = \frac{2\cos^2 2A}{2\sin 2A \cos 2A } -\text{cosec }{8A}\\ \ & = \frac{\cos 2A}{\sin 2A } -\text{cosec }{8A}\\ \ & = \cot 2A -\text{cosec }{8A}\\ \ & =\text{RHS} \end{align}

12. (c) Prove that: \( \frac{\sec4\theta-1}{\sec2\theta-1}=\tan4\theta\cot\theta \)

Solution:

\begin{align}\text{LHS} & =\frac{\sec4\theta-1}{\sec2\theta-1} \\ & = \dfrac{\dfrac{1}{\cos4\theta}-1}{\dfrac{1}{\cos2\theta}-1} \\  & = \dfrac{\dfrac{1-\cos4\theta}{\cos4\theta}}{\dfrac{1-\cos2\theta}{\cos2\theta}} \\ & =\frac{1-\cos4\theta}{\cos4\theta}\times\frac{\cos2\theta}{1-\cos2\theta} \\  & =\frac{1-\left(1-2\sin^22\theta\right)}{\cos4\theta}\times\frac{\cos2\theta}{1-\left(1-2\sin^2\theta\right)} \\  & =\frac{2\sin^22\theta}{\cos4\theta}\times\frac{\cos2\theta}{2\sin^2\theta} \\  & =\frac{2\sin2\theta\cos2\theta}{\cos4\theta}\times\frac{\sin2\theta}{2\sin^2\theta} \\  & =\frac{\sin4\theta}{\cos4\theta}\times\frac{2\sin\theta\cos\theta}{2\sin^2\theta} \\  & =\tan4\theta\times\frac{\cos\theta}{\sin\theta} \\  & =\tan4\theta.\cot\theta \\  & = \text{ RHS} \end{align}

12. (d) Prove that: \( \displaystyle \frac{\sec8A-1}{\sec4A-1}=\frac{\tan8A}{\tan2A} \)

Solution:

\begin{align}\text{LHS } & =\frac{\sec8A-1}{\sec4A-1}\\ & = \dfrac{\dfrac{1}{\cos8A}-1}{\dfrac{1}{\cos4A}-1}\\  & = \dfrac{\dfrac{1-\cos8A}{\cos8A}}{\dfrac{1-\cos4A}{\cos4A}}\\  & =\frac{1-\cos8A}{\cos8A}\times\frac{\cos4A}{1-\cos4A}\\  & =\frac{1-\left(1-2\sin^24A\right)}{\cos8A}\times\frac{\cos4A}{1-\left(1-2\sin^2 2A\right)}\\  & =\frac{2\sin^24A}{\cos8A}\times\frac{\cos4A}{2\sin^2 2A}\\  & =\frac{2\sin4A\cos4A}{\cos8A}\times\frac{\sin4A}{2\sin^2 2A}\\  & =\frac{\sin8A}{\cos8A}\times\frac{2\sin2A\cos2A}{2\sin^2 2A}\\  \ & =\tan8A\times\frac{\cos2A}{\sin2A}\\  & =\tan8A.\cot2A\\ & = \tan 8A \times \frac{1}{\tan 2A}\\ & = \frac{\tan 8A}{\tan 2A} \\  & = \text{ RHS} \end{align}

12. (e) Prove that: \( \tan\theta+2\tan2\theta+4\tan4\theta+8\cot8\theta=\cot\theta \)

Solution:

\begin{align} \text{LHS } &=\tan\theta+2\tan2\theta+4\tan4\theta+8\cot8\theta\\ \ &=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{8}{\tan8\theta}\\ \ &=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{8}{\frac{2\tan4\theta}{1-\tan^24\theta}}\\ \ &=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{8\left(1-\tan^24\theta\right)}{2\tan4\theta}\\ \ &=\tan\theta+2\tan2\theta+4\tan4\theta+\frac{4\left(1-\tan^24\theta\right)}{\tan4\theta}\\ \ &=\tan\theta+2\tan2\theta+\frac{4\tan^24\theta+4-4\tan^24\theta}{\tan4\theta}\\ \ &=\tan\theta+2\tan2\theta+\frac{4}{\tan4\theta}\\ \ &=\tan\theta+2\tan2\theta+\frac{4}{\frac{2\tan2\theta}{1-\tan^22\theta}}\\ \ &=\tan\theta+2\tan2\theta+\frac{4\left(1-\tan^22\theta\right)}{2\tan2\theta}\\ \ &=\tan\theta+2\tan2\theta+\frac{2\left(1-\tan^22\theta\right)}{\tan2\theta}\\ \ &=\tan\theta+2\tan2\theta+\frac{2-2\tan^22\theta}{\tan2\theta}\\ \ &=\tan\theta+\frac{2\tan^22\theta+2-2\tan^22\theta}{\tan2\theta}\\ \ &=\tan\theta+\frac{2}{\tan2\theta}\\ \ &=\tan\theta+\frac{2}{\frac{2\tan\theta}{1-\tan^2\theta}}\\ \ &=\tan\theta+\frac{2\left(1-\tan^2\theta\right)}{2\tan\theta}\\ \ &=\tan\theta+\frac{1-\tan^2\theta}{\tan\theta}\\ \ &=\frac{\tan^2\theta+1-\tan^2\theta}{\tan\theta}\\ \ &=\frac{1}{\tan\theta}\\ \ &=\cot\theta\\ \ &=\text{ RHS} \end{align}

13. (a) Prove that:

\( 4\left(\cos^310^\circ+\sin^320^\circ\right)=3\left(\cos10^\circ+\sin20^\circ\right) \)

Solution:

\begin{align} \text{LHS } \ & =4\left(\cos^310^\circ+\sin^320^\circ\right)\\ \ & =4\cos^310^\circ+4\sin^320^\circ\\ \ & =3\cos10^\circ+\cos3\left(10^\circ\right)+3\sin20^\circ-\sin3\left(20^\circ\right)\\ \ & =3\cos10^\circ+\cos30^\circ+3\sin20^\circ-\sin60^\circ\\ \ & =3\cos10^\circ+\frac{\sqrt{3}}{2}+3\sin20^\circ-\frac{\sqrt{3}}{2}\\ \ & =3\cos10^\circ+3\cos20^\circ\\ \ & =3\left(\cos10^\circ+\cos20^\circ\right)\\ \ & =\text{ RHS } \end{align}

13. (b) Prove that: \( \sin^3\theta\cos3\theta+\cos^3\theta\sin3\theta=\frac{3}{4}\sin4\theta \)

Solution:

\begin{align}\text{LHS } \ & =\sin^3\theta\cos3\theta+\cos^3\theta\sin3\theta\\ \ & =\frac{3\sin\theta-\sin3\theta}{4}\times\cos3\theta+\frac{3\cos\theta+\cos3\theta}{4}\times\sin3\theta\\ \ & =\frac{3\cos3\theta\sin\theta-\sin3\theta\cos3\theta}{4}+\frac{3\sin3\theta\cos\theta+\sin3\theta\cos3\theta}{4}\\ \ & =\frac{3\cos3\theta\sin\theta-\sin3\theta\cos3\theta+3\sin3\theta\cos\theta+\sin3\theta\cos3\theta}{4}\\ \ & =\frac{3\left(\cos3\theta\sin\theta+\sin3\theta\cos\theta\right)}{4}\\ \ & =\frac{3\sin\left(3\theta+\theta\right)}{4}\\ \ & =\frac{3}{4}\sin4\theta\\ \ & =\text{ RHS } \end{align}

13. (c) Prove that:\( \cos^3\theta\cos3\theta+\sin^3\theta\sin3A=\cos^32\theta \)

Solution:

\begin{align} \text{LHS } \ & =\cos^3\theta.\cos3\theta+\sin^3\theta.\sin3\theta\\ \ & =\frac{3\cos\theta+\cos3\theta}{4}\times\cos3\theta+\frac{3\sin\theta-\sin3\theta}{4}\times\sin3\theta\\ \ & =\frac{3\cos3\theta\cos\theta+\cos^23\theta}{4}+\frac{3\sin3\theta\sin\theta-\sin^23\theta}{4}\\ \ & =\frac{3\cos3\theta\cos\theta+\cos^23\theta+3\sin3\theta\sin\theta-\sin^23\theta}{4}\\ \ & =\frac{3\left(\cos3\theta\cos\theta+\sin3\theta\sin\theta\right)+\left(\cos^23\theta-\sin^23\theta\right)}{4}\\ \ & =\frac{3\cos\left(3\theta-\theta\right)+\cos2\left(3\theta\right)}{4}\\ \ & =\frac{3\cos\left(3\theta-\theta\right)+\cos6\theta}{4}\\ \ & =\frac{3\cos2\theta+\cos3\left(2\theta\right)}{4}\\ \ & =\frac{3\cos2\theta+4\cos^3\theta-3\cos\theta}{4}\\ \ & =\frac{4\cos^3\theta}{4}\\ \ & =\cos^3\theta\\ \ & = \text{ RHS} \end{align}

13. (d) Prove that: \( \displaystyle \tan A + \tan \left(\frac{\pi^c}{3} \right) - \tan \left(\frac{\pi^c}{3} \right) \)

Solution:

\begin{align}\text{ LHS} & =\tan A + \tan\left(\frac{\pi^c}{3}+A\right)-\tan \left(\frac{\pi^c}{3}-A\right)\\ \ &=\tan A + \tan\left(60^{\circ}+A\right)-\tan \left(60^{\circ}-A\right)\\ \ & = \tan A + \frac{\tan60^{\circ}+\tan A}{1-\tan 60^{\circ}\tan A} - \frac{\tan60^{\circ}-\tan A}{1+\tan 60^{\circ}\tan A}\\ \ & = \tan A+ \frac{\sqrt{3}+\tan A}{1-\sqrt{3}\tan A}- \frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}\\ \ & = \tan A+ \frac{(\sqrt{3}+\tan A)(1+\sqrt{3}\tan A)-(\sqrt{3}-\tan A)(1-\sqrt{3}\tan A)}{(1-\sqrt{3}\tan A)(1+\sqrt{3}\tan A)}\\ \ & = \tan A+\frac{\sqrt{3}+3\tan A+\tan A+\sqrt{3}\tan^2 A-(\sqrt{3}-3\tan A-\tan A+\sqrt{3}\tan^2 A)}{1^2 -(\sqrt{3}\tan A)^2 } \\ \ & = \tan A + \frac{\sqrt{3}+3\tan A+\tan A+\sqrt{3}\tan^2 A-\sqrt{3}+3\tan A+\tan A-\sqrt{3}\tan^2 A}{1-3\tan^2 A}\\ \ & = \tan A +\frac{8\tan A}{1-3\tan^2 A}\\ \ & = \frac{\tan A(1-3\tan^2 A)+8\tan A}{1-3\tan^2 A}\\ \ & = \frac{\tan A -3\tan^3 A+ 8 \tan A}{1-3\tan^2 A}\\ \ & =\frac{9\tan A -3\tan^3 A}{1-3\tan^2 A}\\ \ & =3\left(\frac{3\tan A -\tan^3 A}{1-3\tan^2 A}\right)\\ \ & = 3\tan 3A \\ \ & = \text{ RHS} \end{align}

14 (a) If \( 2 \tan A = 3\tan B \) prove that \( \tan( A+ B)=\frac{5\sin2 B}{5\cos2 B-1} \)

Solution:

Given,

\begin{align} 2\tan A & =3\tan B \\ or, \tan A & =\frac{3\tan B}{2} \end{align}

Now,

\begin{align}\text{LHS } \ & =\tan\left( A+ B\right)\\ \ & =\frac{\tan A+\tan B}{1-\tan A\tan B}\\ \ & =\frac{\frac{3\tan B}{2}+\tan B}{1-\frac{3\tan B}{2}\times\tan B}\\ \ & =\frac{\frac{3\tan B+2\tan B}{2}}{\frac{2-3\tan^2 B}{2}}\\ \ & =\frac{5\tan B}{2-3\tan^2 B}\\ \ & =\frac{5\frac{\sin B}{\cos B}}{2-3\times\frac{\sin^2 B}{\cos^2 B}}\\ \ & =\frac{5\frac{\sin B}{\cos B}}{\frac{2\cos^2 B-3\sin^2 B}{\cos^2 B}}\\ \ & =\frac{5\sin B}{\cos B}\times\frac{\cos^2 B}{2\cos^2 B-3\sin^2 B}\\ \ & =\frac{5\sin B\cos B}{2\cos^2 B-3\sin^2 B}\times\frac{2}{2}\\ \ & =\frac{10\sin B\cos B}{4\cos^2 B-6\sin^2 B}\\ \ & =\frac{5\left(2\sin B\cos B\right)}{4\times\frac{1+\cos2 B}{2}-6\times\frac{1-\cos2 B}{2}}\\ \ & =\frac{5\sin2 B}{2+2\cos2 B-3+3\cos2 B}\\ \ & =\frac{5\sin2 B}{5\cos2 B-1}\\ \ & = \text{ RHS} \end{align}

14 (b) \( 2 \tan A = 3\tan B \) prove that: \( \tan ( A - B ) = \frac {\sin 2 B }{5-\cos 2 B} \)

Solution:

Given,

\begin{align}2\tan A &=3\tan B \\ \tan A & =\frac{3\tan B}{2} \end{align}

\begin{align} \text{LHS } \ &=\tan\left( A- B\right) \\ \ &=\frac{\tan A-\tan B}{1+\tan A\tan B} \\ \ &=\frac{\frac{3\tan B}{2}-\tan B}{1+\frac{3\tan B}{2}\times\tan B} \\ \ &=\frac{\frac{3\tan B-2\tan B}{2}}{\frac{2-3\tan^2 B}{2}} \\ \ &=\frac{\tan B}{2+3\tan^2 B} \\ \ &=\frac{\frac{\sin B}{\cos B}}{2+3\times\frac{\sin^2 B}{\cos^2 B}} \\ \ &=\frac{\frac{\sin B}{\cos B}}{\frac{2\cos^2 B+3\sin^2 B}{\cos^2 B}} \\ \ &=\frac{\sin B}{\cos B}\times\frac{\cos^2 B}{2\cos^2 B+3\sin^2 B} \\ \ &=\frac{\sin B\cos B}{2\cos^2 B+3\sin^2 B}\times\frac{2}{2} \\ \ &=\frac{2\sin B\cos B}{4\cos^2 B+6\sin^2 B} \\ \ &=\frac{\sin2 B}{4\times\frac{1+\cos2 B}{2}+6\times\frac{1-\cos2 B}{2}} \\ \ &=\frac{\sin2 B}{2+2\cos2 B+3-3\sin2 B} \\ \ &=\frac{\sin2 B}{5-\cos2 B} \\ \ &= \text{ RHS} \end{align}

15. (a) Prove that: \( \sin^4 A=\frac{1}{8}\left(3-4\cos2 A+\cos4 A\right) \)

Solution:

\begin{align} \text{LHS }  \ & =\sin^4 A \\  \ & =\left(\sin^2 A\right)^2 \\  \ & =\left(\frac{1-\cos2 A}{2}\right)^2 \\  \ & =\frac{1}{4}\left(1-\cos2 A\right)^2 \\  \ & =\frac{1}{4}\left(1^2-2\times1\times\cos2 A+\cos^22 A\right) \\  \ & =\frac{1}{4}\left(1-2\cos2 A+\frac{1+\cos4 A}{2}\right) \\  \ & =\frac{1}{4}\left(\frac{2-4\cos2 A+1+\cos4 A}{2}\right) \\  \ & =\frac{3-4\cos2 A+\cos4 A}{8} \\  \ & =\frac{1}{8}(3-4\cos2 A+\cos4 A) \\  \ & = \text{ RHS} \end{align}

15. (b) Prove that: \( \cos^4 A=\frac{1}{8}\left(3+4\cos2 A+\cos4 A\right) \)

Solution:

\begin{align}\text{LHS}  \ &=\cos^4 A \\ \ &=\left(\cos^2 A\right)^2 \\ \ &=\left(\frac{1+\cos2 A}{2}\right)^2 \\  \ &=\frac{1}{4}\left(1+\cos2 A\right)^2 \\ \ &=\frac{1}{4}\left(1^2+2\times1\times\cos2 A+\cos^22 A\right) \\  \ &=\frac{1}{4}\left(1+2\cos2 A+\frac{1+\cos4 A}{2}\right) \\  \ &=\frac{1}{4}\left(\frac{2+4\cos2 A+1+\cos4 A}{2}\right) \\ \ &=\frac{3+4\cos2 A+\cos4 A}{8} \\ \ &= \frac{3}{8}+ \frac{4\cos 2A }{8} + \frac{\cos 4A}{8} \\ \ &= \frac{3}{8}+ \frac{1}{2}\cos 2A + \frac{1}{8}\cos 4A \\  \ &= \text{ RHS} \end{align}

15. (c) Prove that: \( \sin5\theta=16\sin^5\theta-20\sin^3\theta+5\sin \theta \)

Solution:

\begin{align} \text{LHS } \ &=\sin5\theta \\ \ &=\sin\left(3\theta+2\theta\right) \\ \ &=\sin3\theta\cos2\theta+\cos3\theta\sin2\theta \\ \ &=\left(3\sin \theta-4\sin^3\theta\right)\cos2\theta+\left(4\cos^3\theta-3\cos \theta\right)\sin2\theta \\ \ &=3\sin \theta\cos2\theta-4\sin^3\theta\cos2\theta+4\cos^3\theta\sin2\theta-3\cos \theta\sin2\theta \\ \ &=3\sin \theta\left(1-2\sin^2\theta\right)-4\sin^3\theta\left(1-2\sin^2\theta\right) \\ \ &\ \ \ +4\cos^3\theta\left(2\sin \theta\cos \theta\right)-3\cos \theta\left(2\sin \theta\cos \theta\right) \\ \ &=3\sin \theta-6\sin^3\theta-4\sin^3\theta+8\sin^5\theta \\ \ &\ \ \ +8\sin \theta\cos^4\theta-6\sin \theta\cos^2\theta \\ \ &=3\sin \theta-10\sin^3\theta+8\sin^5\theta \\ \ &\ \ \ +8\sin \theta\left(\cos^2\theta\right)^2-6\sin \theta\left(1-\sin^2\theta\right) \\ \ &=3\sin \theta-10\sin^3\theta+8\sin^5\theta+8\sin \theta\left(1-\sin^2\theta\right)^2-6\sin \theta+6\sin^3\theta \\ \ &=8\sin^5\theta-4\sin^3\theta-3\sin \theta+8\sin \theta\left(1-2\sin^2\theta+\sin^4\theta\right) \\ \ &=8\sin^5\theta-4\sin^3\theta-3\sin \theta+8\sin \theta-16\sin^3\theta+8\sin^5\theta \\ \ &=16\sin^5\theta-20\sin^3\theta+5\sin \theta \\ \ &= \text{ RHS} \end{align}

15 (d) Prove that: \( \cos5A=16\cos^5A-20\cos^3A+5\cos A \)

Solution:

\begin{align} \text{LHS } \ & =\cos5A \\ \ & =\cos\left(3A+2A\right) \\ \ & =\cos3A\cos2A-\sin3A\sin2A \\ \ & =\left(4\cos^3A-3\cos A\right)\cos2A \\ \ &\ \ \ \ -\left(3\sin A-4\sin^3A\right)\sin2A \\ \ & =4\cos^3A\cos2A-3\cos A\cos2A \\ \ & \ \ \ \ -3\text{sinA}\sin2A+4\sin^3A\sin2A \\ \ & =4\cos^3A\left(2\cos^2A-1\right)-3\cos A\left(2\cos^2A-1\right) \\ \ &\ \ \ \ -3\sin A\left(2\sin A\cos A\right)+4\sin^3A\left(2\sin A\cos A\right) \\ \ & =8\cos^5A-4\cos^3A-6\cos^3A+3\cos A \\ \ & \ \ \ \ -6\sin^2A\cos A+8\sin^4A\cos A \\ \ & =8\cos^5A-10\cos^3A+3\cos A \\ \ & \ \ \ \ -6\left(1-\cos^2A\right)\cos A+8\left(\sin^2A\right)^2\cos A \\ \ & =8\cos^5A-10\cos^3A+3\cos A \\ \ & \ \ \ \ -6\cos A+6\cos^3A+\cdot8\left(1-\cos^2A\right)^2\cos A \\ \ & =8\cos^5A-4\cos^3A-3\cos A \\ \ & \ \ \ \ +8\left(1-2\cos^2A+\cos^4A\right)\cos A \\ \ & =8\cos^5A-4\cos^3A-3\cos A \\ \ & \ \ \ \ +8\cos A-16\cos^3A+8\cos^5A \\ \ & =16\cos^5A-20\cos^3A+5\cos A \\ \ & = \text{ RHS} \end{align}

16. With the help of multiple angles relation of Sine and Cosine, find the value of \( \sin18^{\circ}, \sin36^{\circ} \) and \( \sin54^{\circ} \). By using these values, find the values of \( \cos18^{\circ}, \cos36^{\circ} \) and \( \cos54^{\circ} \). Also, find the value of \( \tan18^{\circ}, \tan36^{\circ} \) and \( \tan54^{\circ} \) . Share your result to your friend and prepare combine report.

Solution:

\begin{align} \text{ Let, } & \theta = 18^{\circ} \\ or, & \ 5\theta = 5\times 18^{\circ} \\ or, & \ 2\theta + 3\theta = 90^{\circ} \\ or, & \ 3\theta = 90^{\circ} - 2\theta \\ or, & \ \cos 3\theta = \cos (90^{\circ} - 2\theta )\\ or, & \ 4\cos^3\theta -3\cos \theta = \sin 2\theta \\ or, & \ \cos \theta [4\cos^2 \theta - 3 ] = 2\sin \theta \cos \theta \\ or, & \ 4(1-\sin^2 \theta ) -3 = 2\sin \theta \\ or, & \ 4 -4\sin^2 \theta -3 -2\sin \theta =0 \\ or, & \ -4\sin^2 \theta -2\sin \theta +1 = 0 \\ or, & \ -(4\sin^2 \theta +2\sin \theta -1) =0 \\ or, & \ 4\sin^2 \theta +2\sin \theta -1=0 \end{align}

Now, comparing with \( ax^2+bx+c=0 \), we get

\( a=4, b= 2, c= -1, x = \sin \theta \)

Now,

\begin{align} \sin \theta & = \frac{-b\pm \sqrt{b^2 - 4ac}}{2a} \\ \ & =\frac{-2\pm \sqrt{(2)^2 -4 \times 4 \times (-1)}}{2\times 4 } \\ \ & = \frac{-2\pm \sqrt{4+16}}{8} \\ \ & = \frac{-2\pm\sqrt{20}}{8} \\ \ & = \frac{-2\pm 2\sqrt{5}}{8} \\ \ & = \frac{2(-1\pm \sqrt{5})}{8}\\ \ & =\frac{-1\pm\sqrt{5}}{4} \end{align}

Since \( \sin 18^{\circ} \) is positive,

\( \therefore \sin 18^{\circ} = \frac{-1+\sqrt{5}}{4}=\frac{\sqrt{5}-1}{4} \)

Now,

\begin{align} \cos 18^{\circ} &=\sqrt{1-\sin^2 18^{\circ}}\\ \ & = \sqrt{1-\left(\frac{\sqrt{5}-1}{4}\right)^2}\\ \ & = \sqrt{1-\frac{\left\{(\sqrt{5})^2-2\times \sqrt{5}\times 1+1^2 \right\}^2}{16}}\\ \ & = \sqrt{\frac{16-(5-2\sqrt{5}+1)}{16}}\\ \ & = \sqrt{\frac{16-5+2\sqrt{5}-1}{16}}\\ \ & = \sqrt{\frac{10+2\sqrt{5}}{16}}\\ \ & = \frac{\sqrt{10+2\sqrt{5}}}{4}\\ \therefore \cos 18^{\circ}& = \frac{\sqrt{10+2\sqrt{5}}}{4} \end{align}

Now,

\begin{align} \tan 18^{\circ}& =\frac{\sin 18^{\circ}}{\cos 18^{\circ}}\\ \ & = \frac{\frac{\sqrt{5}-1}{4}}{\frac{\sqrt{10+2\sqrt{5}}}{4}}\\ \ & = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\\ \therefore \tan 18^{\circ}& = \frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}\\ \end{align}

\begin{align} \cos 36^{\circ}& = 1-2\sin^2 18^{\circ}\\ \ & = 1-2\left(\frac{\sqrt{5}-1}{4} \right)^2\\ \ & = 1 - 2\times \frac{(\sqrt{5})^2-2\times \sqrt{5}\times 1 +1^2 }{16}\\ \ & = 1 - \frac{5-2\sqrt{5}+1}{8} \\ \ & = \frac{8-5+2\sqrt{5}-1}{8} \\ \ & = \frac{2+2\sqrt{5}}{8} \\ \ & = \frac{2(1+\sqrt{5})}{8} \\ \text{or, } \cos 36^{\circ} &= \frac{1+\sqrt{5}}{4}\\ \therefore \cos 36^{\circ}& = \frac{\sqrt{5}+1}{4}\\ \end{align}

Also,

\begin{align} \text{We know,}\\ \sin 36^{\circ}& = \sqrt{1-\cos^2 36^{\circ}}\\ \ & = \sqrt{1-\left( \frac{\sqrt{5}+1}{4} \right)^2}\\ \ & = \sqrt{1-\frac{(\sqrt{5})^2 +2\times \sqrt{5}\times 1+1^2 }{16}} \\ \ & = \sqrt{1-\frac{5 +2\times \sqrt{5}\times 1+1 }{16}} \\ \ & = \sqrt{1-\frac{6 + 2\sqrt{5} }{16}} \\ \ & = \sqrt{\frac{16-6 - 2\sqrt{5} }{16}} \\ \ & = \sqrt{\frac{10- 2\sqrt{5} }{16}} \\ \ & = \frac{\sqrt{10-2\sqrt{5}}}{4}\\ \therefore \sin 36^{\circ}\ & = \frac{\sqrt{10-2\sqrt{5}}}{4}\\ \end{align}

Also,

\begin{align} \tan 36^{\circ} & = \frac{\sin 36^{\circ} }{\cos 36^{\circ} }\\ \ & = \frac{ \frac{\sqrt{10-2\sqrt{5}}}{4}\\}{\frac{\sqrt{5}+1}{4}}\\ \ & =\frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}+1}\\ \therefore \tan 36^{\circ}& = \frac{\sqrt{10-2\sqrt{5}}}{\sqrt{5}+1}\\ \end{align}

Also

\begin{align} \sin 54^{\circ} & = \sin(90^{\circ}-36^{\circ})\\ \ & =\cos 36^{\circ}\\ \ & = \frac{\sqrt{5}+1}{4}\\ \therefore \sin 54^{\circ}&=\frac{\sqrt{5}+1}{4}\\ \end{align}

Also,

\begin{align} \cos 54^{\circ} & = \cos(90^{\circ}-36^{\circ})\\ \ & =\sin 36^{\circ}\\ \ & = \frac{\sqrt{10-2\sqrt{5}}}{4}\\ \therefore \cos 54^{\circ}&=\frac{\sqrt{10-2\sqrt{5}}}{4}\\ \end{align}

Also,

\begin{align} \tan 54^{\circ} & = \frac{\sin 54^{\circ} } {\cos 54^{\circ} }\\ \ & =\frac{\frac{\sqrt{5}+1}{4} } { \frac{\sqrt{10-2\sqrt{5}}}{4}} \\ \ & =\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}} \\ \therefore \ \tan54^{\circ} & =\frac{\sqrt{5}+1}{\sqrt{10-2\sqrt{5}}} \\ \end{align}

All The Best