Conditional Trigonometric Identities
The trigonometric identities which are true only under certain conditions are known as conditional trigonometric identities. Examples of conditional identities are: (i) If \(A + B + C = \pi^c\) then \( \sin (A + B) = \sin C\) (ii) If \( A + B + C = \pi^c\) then \( \tan (A + B) = – \tan C\) (iii) If \( A + B = 90^{\circ}\) then \( \sin A = cos B\)
1. (b) What is the true condition for the identity \( \tan A = \cot B\) ?
\( \tan A = \cot B\) is true, only when \( A+ B = \frac{\pi^c}{2}\)
1. (c) Write any three relations which can be formed from \(A + B + C = \pi^c\)
Any three relations that can be formed from \(A + B + C = \pi^c\) are \begin{align} \ & (i) \ A+B = \pi^c - C \\ \ & (ii) \ \frac{A}{2} + \frac{B}{2} = \frac{\pi^c}{2} - \frac{C}{2} \\ \ & (iii) \ 2A+2B = 2\pi^c - 2C \\ \end{align}
2. (a) If \( A+B+C = \pi^c\), prove that: \( \tan A + \tan B + \tan C = \tan A \tan B \tan C \)
\begin{align} \text{ Here }, \ & \\ \ & A+B+C =\pi^c\\ or, \ & A+B =\pi^c -C \\ \ & \text{Taking tan on both sides }\\ or, \ & \tan (A+B) =\tan(\pi^c -C)\\ or, \ & \frac{\tan A + \tan B }{1- \tan A \tan B} = -\tan C\\ or, \ & \tan A + \tan B = -\tan C( 1-\tan A \tan B)\\ or, \ & \tan A + \tan B = -\tan C + \tan A \tan B \tan C\\ \therefore \ & \tan A +\tan B + \tan C =\tan A \tan B \tan C \\ \end{align}
2. (b) If \( A+B+C = \pi^c\), prove that: \( \tan \frac{A}{2} \tan \frac{B}{2} +\tan \frac{B}{2} \tan \frac{C}{2} +\tan \frac{C}{2} \tan \frac{A}{2} =1\)
\begin{align} \text{ Here }, \ & \\ \ & A+B+C =\pi^c\\ or, \ & \frac{A}{2} + \frac{B}{2} +\frac{C}{2}=\frac{ \pi^c}{2}\\ or, \ & \frac{A}{2} + \frac{B}{2} =\frac{ \pi^c}{2}-\frac{C}{2}\\ \ & \text{Taking tan on both sides }\\ or, \ & \tan \left(\frac{A}{2} + \frac{B}{2}\right) =\tan \left(\frac{ \pi^c}{2}-\frac{C}{2}\right)\\ or, \ & \frac{\tan \frac{A}{2} + \tan \frac{B }{2}}{1- \tan \frac{A}{2} \tan \frac{B}{2}} = \cot \frac{C}{2} \\ or, \ & \frac{\tan \frac{A}{2} + \tan \frac{B }{2}}{1- \tan \frac{A}{2} \tan \frac{B}{2}} = \frac{1}{\tan \frac{C}{2} }\\ or, \ & \tan \frac{C}{2}\left(\tan \frac{A}{2} + \tan \frac{B }{2} \right)= 1- \tan \frac{A}{2} \tan \frac{B}{2} \\ or, \ & \tan \frac{C}{2} \tan \frac{A}{2} + \tan \frac{B}{2} \tan \frac{C}{2} = 1- \tan \frac{A}{2} \tan \frac{B}{2} \\ \therefore \ & \tan \frac{A}{2} \tan \frac{B}{2} +\tan \frac{B}{2} \tan \frac{C}{2} +\tan \frac{C}{2} \tan \frac{A}{2} =1 \\ \end{align}
2. (c) If \( A+B+C = \pi^c\), prove that: \( \tan 2A + \tan 2B + \tan 2C = \tan 2A \tan 2B \tan 2C \)
2. (d) If \( A+B+C = \pi^c\), prove that: \( \cot A \cot B + \cot B \cot C + \cot C \cot A -1 =0 \)
\begin{align} \text{ Here }, \ & \\ \ & A+B+C =\pi^c\\ or, \ & A+B =\pi^c -C \\ \ & \text{Taking cot on both sides }\\ or, \ & \cot (A+B) =\cot (\pi^c -C)\\ or, \ & \frac{\cot A \cot B -1}{\cot A + \cot B} = -\cot C\\ or, \ & \cot A \cot B -1 = -\cot C( \cot A + \cot B)\\ or, \ & \cot A \cot B -1 = -\cot C \cot A - \cot B \cot C\\ \therefore \ & \cot A \cot B + \cot B \cot C + \cot C \cot A -1 =0 \\ \end{align}
2. (e) If \( A+B+C = \pi^c\), prove that: \( \cot 2A \cot 2B + \cot 2B \cot2C + \cot 2C \cot 2A = 1 \)
3. (a) If \( A, B, \text{ and } C\) are the vertices of \( \triangle ABC\) then prove that: \( \sin A + \sin B + \sin C = 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}\)
\begin{align} \text{Given } \ & \\ \ & A+B+C = 180^{\circ}\\ \text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\ \text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\ \text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\ \text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\ \text{Now} \ & \\ \text{LHS} \ & = \sin A + \sin B + \sin C\\ \ & = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + \sin C\\ \ & = 2 \sin \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} + \frac{B}{2} \right) + 2\sin \frac{C}{2}\cos \frac{C}{2} \\ \ & = 2\cos \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) + 2\sin \frac{C}{2}\cos \frac{C}{2} \\ \ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \sin \frac{C}{2} \right] \\ \ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] \\ \ & = 2\cos \frac{C}{2} \left[ 2\cos \frac{A}{2} \cos \frac{B}{2}\right] \\ \ & = 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\ \ & = \text{RHS}\\ \end{align}
3. (b) If \( A, B, \text{ and } C\) are the vertices of \( \triangle ABC\) then prove that: \( \sin A + \sin B - \sin C = 4\sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}\)
\begin{align} \text{Given } \ & \\ \ & A+B+C = 180^{\circ}\\ \text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\ \text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\ \text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\ \text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\ \text{Now} \ & \\ \text{LHS} \ & = \sin A + \sin B - \sin C\\ \ & = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} - \sin C\\ \ & = 2 \sin \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} + \frac{B}{2} \right) -2\sin \frac{C}{2}\cos \frac{C}{2} \\ \ & = 2\cos \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) - 2\sin \frac{C}{2}\cos \frac{C}{2} \\ \ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) - \sin \frac{C}{2} \right] \\ \ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) - \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] \\ \ & = 2\cos \frac{C}{2} \left[ 2\sin \frac{A}{2} \sin \frac{B}{2}\right] \\ \ & = 4\sin \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}\\ \ & = \text{RHS}\\ \end{align}
3. (c) If \( A, B, \text{ and } C\) are the vertices of \( \triangle ABC\) then prove that: \( \sin A - \sin B - \sin C = -4\cos \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\)
\begin{align} \text{Given } \ & \\ \ & A+B+C = 180^{\circ}\\ \text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\ \text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\ \text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\ \text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\ \text{Now} \ & \\ \text{LHS} \ & = \sin A - \sin B - \sin C\\ \ & = 2\cos \frac{A+B}{2} \sin \frac{A-B}{2} - \sin C\\ \ & = 2 \cos \left( \frac{A}{2} + \frac{B}{2} \right) \sin \left( \frac{A}{2} + \frac{B}{2} \right) -2\sin \frac{C}{2}\cos \frac{C}{2} \\ \ & = 2\sin \frac{C}{2} \sin \left( \frac{A}{2} - \frac{B}{2} \right) - 2\sin \frac{C}{2}\cos \frac{C}{2} \\ \ & = 2\sin \frac{C}{2} \left[ \sin \left( \frac{A}{2} - \frac{B}{2} \right) - \cos \frac{C}{2} \right] \\ \ & = 2\sin \frac{C}{2} \left[ \sin \left( \frac{A}{2} - \frac{B}{2} \right) - \sin \left( \frac{A}{2} + \frac{B}{2} \right) \right] \\ \ & = - 2\sin \frac{C}{2} \left[ \sin \left( \frac{A}{2} + \frac{B}{2} \right) -\sin \left( \frac{A}{2} - \frac{B}{2} \right) \right] \\ \ & = -2\sin \frac{C}{2} \left[ 2\cos \frac{A}{2} \sin \frac{B}{2}\right] \\ \ & = - 4\cos \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\\ \ & = \text{RHS}\\ \end{align}
3. (d) If \( A, B, \text{ and } C\) are the vertices of \( \triangle ABC\) then prove that: \( \cos A + \cos B - \cos C =-1+ 4\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}\)
\begin{align} \text{Given } \ & \\ \ & A+B+C = 180^{\circ}\\ \text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\ \text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\ \text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\ \text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\ \text{Now} \ & \\ \text{LHS} \ & = \cos A + \cos B - \cos C\\ \ & = 2\cos \frac{A+B}{2} \cos \frac{A-B}{2} - \cos C\\ \ & = 2 \cos \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} - \frac{B}{2} \right) -\left ( 1-2\sin^2 \frac{C}{2} \right) \\ \ & = 2\sin \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) -1 + 2\sin^2 \frac{C}{2} \\ \ & = 2\sin \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) + 2\sin^2 \frac{C}{2} -1 \\ \ & = 2\sin \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \sin \frac{C}{2} \right] -1 \\ \ & = 2\sin \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] -1\\ \ & = 2\sin \frac{C}{2} \left[ \cos \left( \frac{A}{2} + \frac{B}{2} \right) + \cos \left( \frac{A}{2} - \frac{B}{2} \right) \right] -1 \\ \ & = 2\sin \frac{C}{2} \left[ 2\cos \frac{A}{2} \cos \frac{B}{2}\right] -1 \\ \ & = 4\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} -1 \\ \ & = -1+ 4\cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} \\ \ & = \text{RHS}\\ \end{align}
3. (e) If \( A, B, \text{ and } C\) are the vertices of \( \triangle ABC\) then prove that: \( -\cos A + \cos B + \cos C = 4\sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} - 1\)
\begin{align} \text{Given } \ & \\ \ & A+B+C = 180^{\circ}\\ \text{or, } \ & \frac{A+B+C }{2} =\frac{180^{\circ}}{2}\\ \text{or, } \ & \frac{A}{2} + \frac{B}{2} = 90^{\circ}- \frac{C}{2}\\ \text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(90^{\circ}- \frac{C}{2} \right) = \cos \frac{C}{2}\\ \text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(90^{\circ}- \frac{C}{2} \right) = \sin \frac{C}{2}\\ \text{Now} \ & \\ \text{LHS} \ & = - \cos A + \cos B + \cos C\\ \ & = \cos B - \cos A + \cos C\\ \ & = 2\sin \frac{A+B}{2} \sin \frac{A-B}{2} + \cos C\\ \ & = 2 \sin \left( \frac{A}{2} + \frac{B}{2} \right) \sin \left( \frac{A}{2} - \frac{B}{2} \right) + \left (2\cos^2 \frac{C}{2} -1 \right) \\ \ & = 2\cos \frac{C}{2} \sin \left( \frac{A}{2} - \frac{B}{2} \right) + 2\cos^2 \frac{C}{2} -1 \\ \ & = 2\cos \frac{C}{2} \left[ \sin \left( \frac{A}{2} - \frac{B}{2} \right) + \cos \frac{C}{2} \right] -1 \\ \ & = 2\cos \frac{C}{2} \left[ \sin \left( \frac{A}{2} - \frac{B}{2} \right) + \sin \left( \frac{A}{2} + \frac{B}{2} \right) \right] - 1 \\ \ & = 2\cos \frac{C}{2} \left[ \sin \left( \frac{A}{2} + \frac{B}{2} \right) + \sin \left( \frac{A}{2} - \frac{B}{2} \right) \right] -1 \\ \ & = 2\cos \frac{C}{2} \left[ 2\sin \frac{A}{2} \cos \frac{B}{2}\right] -1 \\ \ & = 4\sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}-1 \\ \ & = \text{RHS}\\ \end{align}
4. (a) If \( A+B+C = \pi^c\), prove that: \( \sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C \)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & A+B+C = 2\pi^c \\ \text{or, } \ & \sin (A+B) = \sin(\pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now,} \ & \\ \text{LHS} \ & = \sin 2A + \sin 2B + \sin 2C \\ \ & = 2\sin \frac{2A+2B}{2} \cos \frac{2A-2B}{2} +\sin 2C\\ \ & = 2\sin (A+B) \cos (A-B) + \sin 2C\\ \ & = 2\sin C \cos (A-B) + 2\sin C \cos C\\ \ & = 2\sin C [ \cos (A-B) + \cos C ]\\ \ & = 2\sin C[ \cos (A-B) -\cos (A+B)]\\ \ & =2\sin C[ 2\sin A \sin B ]\\ \ & = 4\sin A \sin B \sin C \\ \ & = \text{RHS}\\ \end{align}
4. (b) If \( A+B+C = \pi^c\), prove that: \( \sin 2A - \sin 2B + \sin 2C = 4\cos A \sin B \cos C \)
4. (c) If \( A+B+C = \pi^c\), prove that: \( \sin 2A - \sin 2B - \sin 2C = - 4\sin A \cos B \cos C \)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & A+B+C = 2\pi^c \\ \text{or, } \ & \sin (A+B) = \sin(\pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now,}\ & \\ \text{LHS} \ & = \sin 2A - \sin 2B - \sin 2C \\ \ & = 2\cos \frac{2A+2B}{2} \sin\frac{2A-2B}{2} -\sin 2C\\ \ & = 2\cos (A+B) \sin (A-B) - \sin 2C\\ \ & = -2\cos C \sin (A-B) - 2\sin C \cos C\\ \ & = -2\cos C [ \sin (A-B) + \sin C ]\\ \ & = -2\cos C [ \sin C + \sin (A-B) ]\\ \ & = -2\cos C[ \sin (A+B) + \sin (A-B)]\\ \ & =-2\cos C[ 2\sin A \cos B ]\\ \ & = -4\sin A \cos B \cos C \\ \ & = \text{RHS}\\ \end{align}
4. (d) If \( A+B+C = \pi^c\), prove that: \( \cos 2A - \cos 2B + \cos 2C =1 - 4\sin A \cos B \sin C \)
4. (e) If \( A+B+C = \pi^c\), prove that: \( \cos 2A + \cos 2B - \cos 2C =1 - 4\sin A \sin B \cos C \)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & A+B+C = 2\pi^c \\ \text{or, } \ & \sin (A+B) = \sin(\pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now,} \ & \\ \text{LHS} \ & = \cos 2A + \cos 2B - \cos 2C \\ \ & = 2\cos \frac{2A+2B}{2} \cos \frac{2A-2B}{2} -( 2\cos^2 C -1)\\ \ & = 2\cos (A+B) \cos (A-B) - 2\cos^2 C + 1\\ \ & = -2\cos C \cos (A-B) - 2\cos^2 C +1\\ \ & = -2\cos C [ \cos (A-B) + \cos C ] + 1\\ \ & = -2\cos C [ \cos (A-B) - \cos (A+B) ] + 1\\ \ & =-2\cos C[ 2\sin A \sin B ] + 1 \\ \ & = -4\sin A \sin B \cos C +1 \\ \ & = 1-4\sin A \sin B \cos C \\ \ & = \text{RHS}\\ \end{align}
5. (a) If \( A+B+C = \pi^c\), prove that: \( \sin (B+C-A) + \sin (C+A-B) + \sin (A+B-C) =4\sin A \sin B \sin C \)
\begin{align} \ & A+B+C=\pi^c \\ \ & A+B =\pi^c-C ...(i)\\ \ & B+C = \pi^c -A ...(ii)\\ \ & C+A =\pi^c - B ...(iii) \\ \text{Now, }\ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & =\sin (B+C-A) + \sin (C+A-B) + \sin (A+B-C) \\ \ & = \sin (\pi^c -A-A) + \sin (\pi^c - B - B) + \sin (\pi^c-C -C) \\ \ & = \sin (\pi^c -2A) + \sin (\pi^c - 2B) + \sin (\pi^c-2C) \\ \ & = \sin 2A + \sin 2B + \sin 2C \\ \ & = 2\sin \frac{2A+2B}{2} \cos \frac{2A-2B}{2} +\sin 2C\\ \ & = 2\sin (A+B) \cos (A-B) + \sin 2C\\ \ & = 2\sin C \cos (A-B) + 2\sin C \cos C\\ \ & = 2\sin C [ \cos (A-B) + \cos C ]\\ \ & = 2\sin C[ \cos (A-B) -\cos (A+B)]\\ \ & =2\sin C[ 2\sin A \sin B ]\\ \ & = 4\sin A \sin B \sin C \\ \ & = \text{RHS}\\ \end{align}
5. (b) If \( A+B+C = \pi^c\), prove that: \( \cos (B+C-A) + \cos (C+A-B) + \cos (A+B-C) =4\cos A \cos B \cos C +1 \)
\begin{align} \ & A+B+C=\pi^c \\ \ & A+B =\pi^c-C ...(i)\\ \ & B+C = \pi^c -A ...(ii)\\ \ & C+A =\pi^c - B ...(iii) \\ \text{Now, }\ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & =\cos (B+C-A) + \cos (C+A-B) + \cos (A+B-C) \\ \ & = \cos (\pi^c -A-A) + \cos (\pi^c - B - B) + \cos (\pi^c-C -C) \\ \ & = \cos (\pi^c -2A) + \cos (\pi^c - 2B) + \cos (\pi^c-2C) \\ \ & = - \cos 2A - \cos 2B - \cos 2C \\ \ & = - (\cos 2A + \cos 2B) - \cos 2C \\ \ & = -2 \cos \frac{2A+2B}{2} \cos \frac{2A-2B}{2} - \cos 2C \\ \ & =- 2\cos (A+B) \cos (A-B) - \cos 2C \\ \ & = 2\cos C \cos (A-B) - (2\cos^2 C -1 )\\ \ & = 2\cos C \cos (A-B) - 2\cos^2 C + 1 \\ \ & = 2\cos C [ \cos (A-B) - \cos C ]+ 1\\ \ & = 2\cos C[ \cos (A-B) + \cos (A+B)]+ 1 \\ \ & =2\cos C[ 2\cos A \cos B ] + 1 \\ \ & = 4\cos A \cos B \cos C + 1 \\ \ & = \text{RHS}\\ \end{align}
5. (c) If \( A+B+C = \pi^c\), prove that: \( \frac{\cos A}{\sin B \sin C} +\frac{\cos B }{\sin C \sin A } + \frac{\cos C}{\sin B \sin B} =2 \)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & = \frac{\cos A}{\sin B \sin C} +\frac{\cos B }{\sin C \sin A } + \frac{\cos C}{\sin B \sin B}\\ \ & = \frac{ \sin A \cos A + \sin B \cos B + \sin C \cos C}{\sin A \sin B \sin C }\\ \ & = \frac{ \sin A \cos A + \sin B \cos B + \sin C \cos C}{\sin A \sin B \sin C }\times \frac{2}{2} \\ \ & = \frac{ 2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C}{2\sin A \sin B \sin C }\\ \ & = \frac{\sin 2A + \sin 2B + \sin 2C}{ 2\sin A \sin B \sin C}\\ \ & = \frac{2\sin \frac{2A+2B}{2} \cos \frac{2A-2B}{2} +\sin 2C }{ 2\sin A \sin B \sin C}\\ \ & = \frac{2\sin (A+B) \cos (A-B) + \sin 2C}{ 2\sin A \sin B \sin C}\\ \ & = \frac{ 2\sin C \cos (A-B) + 2\sin C \cos C}{2\sin A \sin B \sin C}\\ \ & = \frac{ 2\sin C [ \cos (A-B) + \cos C ] }{2\sin A \sin B \sin C}\\ \ & = \frac{ 2\sin C [ \cos (A-B) - \cos (A+B)] }{2\sin A \sin B \sin C}\\ \ & = \frac{ 2\sin C [ 2\sin A \sin B ] }{2\sin A \sin B \sin C}\\ \ & = 2 \\ \ & = \text{RHS}\\ \end{align}
5. (d) If \( A+B+C = \pi^c\), prove that: \( \frac{\sin A}{\cos B \cos C}+\frac{\sin B}{\cos C \cos A} +\frac{\cos C}{\cos A \cos B } = 2\tan A \tan B \tan C\)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & = \frac{\sin A}{\cos B \cos C}+\frac{\sin B}{\cos C \cos A} +\frac{\cos C}{\cos A \cos B }\\ \ & = \frac{ \sin A \cos A + \sin B \cos B + \sin C \cos C}{\cos A \cos B \cos C }\\ \ & = \frac{ \sin A \cos A + \sin B \cos B + \sin C \cos C}{\cos A \cos B \cos C}\times \frac{2}{2} \\ \ & = \frac{ 2\sin A \cos A + 2\sin B \cos B + 2\sin C \cos C}{2 \cos A \cos B \cos C}\\ \ & = \frac{\sin 2A + \sin 2B + \sin 2C}{ 2\cos A \cos B \cos C}\\ \ & = \frac{2\sin \frac{2A+2B}{2} \cos \frac{2A-2B}{2} +\sin 2C }{ 2\cos A \cos B \cos C}\\ \ & = \frac{2\sin (A+B) \cos (A-B) + \sin 2C}{ 2\cos A \cos B \cos C}\\ \ & = \frac{ 2\sin C \cos (A-B) + 2\sin C \cos C}{2\cos A \cos B \cos C}\\ \ & = \frac{ 2\sin C [ \cos (A-B) + \cos C ] }{2\cos A \cos B \cos C}\\ \ & = \frac{ 2\sin C [ \cos (A-B) - \cos (A+B)] }{2\cos A \cos B \cos C}\\ \ & = \frac{ 2\sin C [ 2\sin A \sin B ] }{2\cos A \cos B \cos C}\\ \ & = 2\frac{ \sin A }{\cos A} \frac{\sin B}{\cos B} \frac{\sin C}{\cos C}\\ \ & = 2\tan A \tan B \tan C \\ \ & = \text{RHS}\\ \end{align}
6. (a) If \( A+B+C = \pi^c\), prove that: \(\cos^2 A + \cos^2 B +\cos^2 C = 1-2\cos A \cos B \cos C\)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & = \cos^2 A + \cos^2 B +\cos^2 C\\ \ & = \frac{1+\cos 2A}{2} + \frac{1+\cos 2B}{2} + \cos^2 C\\ \ & = \frac{1+ \cos 2A + 1+ \cos 2B}{2} + \cos^2 C\\ \ & = \frac{ 2+ \cos 2A + \cos 2B}{2} + \cos^2 C\\ \ & = \frac{ 2+ 2\cos \frac{2A+2B}{2} \cos \frac{2A -2B}{2} }{2} + \cos^2 C\\ \ & = \frac{2+ 2\cos (A+B)\cos (A-B)}{2} + \cos^2 C\\ \ & = 1+ \cos (A+B) \cos (A-B) + \cos^2 C\\ \ & = 1 - \cos C \cos (A-B) + \cos^2 C \\ \ & = 1- \cos C [ \cos (A-B) - \cos C ]\\ \ & = 1- \cos C [ \cos (A-B) + \cos (A+B) ] \\ \ & = 1 - \cos C [2\cos A \cos B ]\\ \ & = 1 - 2\cos A \cos B \cos C \\ \ & = \text{RHS}\\ \end{align}
6. (b) If \( A+B+C = \pi^c\), prove that: \(\cos^2 A + \cos^2 B -\sin^2 C = -2\cos A \cos B \cos C\)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & = \cos^2 A + \cos^2 B - \sin^2 C\\ \ & = \frac{1+\cos 2A}{2} + \frac{1+\cos 2B}{2} - \sin^2 C\\ \ & = \frac{1+ \cos 2A + 1+ \cos 2B}{2} - \sin^2 C\\ \ & = \frac{ 2+ \cos 2A + \cos 2B}{2} - \sin^2 C\\ \ & = \frac{ 2+ 2\cos \frac{2A+2B}{2} \cos \frac{2A -2B}{2} }{2} - ( 1 - \cos^2 C )\\ \ & = \frac{2+ 2\cos (A+B)\cos (A-B)}{2} - 1 + \cos^2 C \\ \ & = 1 + \cos (A+B) \cos (A-B) - 1 + \cos^2 C \\ & = \cos (A+B) \cos (A-B) + \cos^2 C\\ \ & = - \cos C \cos (A-B) + \cos^2 C \\ \ & = - \cos C [ \cos (A-B) - \cos C ]\\ \ & = - \cos C [ \cos (A-B) + \cos (A+B) ] \\ \ & = - \cos C [2\cos A \cos B ]\\ \ & = - 2\cos A \cos B \cos C \\ \ & = \text{RHS}\\ \end{align}
6. (c) If \( A+B+C = \pi^c\), prove that: \(\sin^2 A + \sin^2 B + \sin^2 C = 2(1+ \cos A \cos B \cos C )\)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & = \sin^2 A + \sin^2 B + \sin^2 C \\ \ & = \frac{1- \cos 2A}{2} + \frac{1- \cos 2B}{2} + \sin^2 C\\ \ & = \frac{1- \cos 2A + 1- \cos 2B}{2} + \sin^2 C\\ \ & = \frac{ 2 -( \cos 2A + \cos 2B)}{2} + \sin^2 C\\ \ & = \frac{ 2 - 2\cos \frac{2A+2B}{2} \cos \frac{2A -2B}{2} }{2} + ( 1 - \cos^2 C )\\ \ & = \frac{2- 2\cos (A+B)\cos (A-B)}{2} + 1 - \cos^2 C \\ \ & = 1- \cos (A+B) \cos (A-B) + 1 - \cos^2 C )\\ \ & = 2 + \cos C \cos (A-B) - \cos^2 C \\ \ & = 2+ \cos C [ \cos (A-B) - \cos C ]\\ \ & = 2+ \cos C [ \cos (A-B) + \cos (A+B) ] \\ \ & = 2+ \cos C [2\cos A \cos B ]\\ \ & = 2+ 2\cos A \cos B \cos C \\ \ & = 2( 1+ \cos A \cos B \cos C ) \\ \ & = \text{RHS}\\ \end{align}
6. (d) If \( A+B+C = \pi^c\), prove that: \(\sin^2 A - \sin^2 B - \sin^2 C = - \cos A \sin B \sin C \)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & = \sin^2 A - \sin^2 B - \sin^2 C \\ \ & = \frac{1- \cos 2A}{2} - \frac{1- \cos 2B}{2} - \sin^2 C\\ \ & = \frac{1- \cos 2A - (1- \cos 2B)}{2} - \sin^2 C\\ \ & = \frac{1- \cos 2A - 1+\cos 2B}{2} - \sin^2 C\\ \ & = \frac{ \cos 2B - \cos 2A}{2} - \sin^2 C\\ \ & = \frac{2\sin \frac{2A+2B}{2} \sin \frac{2A-2B}{2} }{2} - \sin^2 C\\ \ & = \frac{2\sin (A+B) \sin (A-B) }{2} - \sin^2 C \\ \ & = \sin (A+B) \sin (A-B) - \sin^2 C \\ \ & = \sin C \sin (A-B) - \sin^2 C\\ \ & = \sin C [ \sin (A-B) - \sin C ]\\ \ & = \sin C [ \sin (A - B) - \sin (A+B) ]\\ \ & = - \sin C [ \sin (A + B) - \sin (A - B) ]\\ \ & = - \sin C [ 2\cos A \sin B ]\\ \ & = -2\cos A \sin B \sin C \\ \ & = \text{RHS}\\ \end{align}
6. (e) If \( A+B+C = \pi^c\), prove that: \(\sin^2 A - \sin^2 B + \sin^2 C = \sin A \cos B \sin C \)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & = \sin^2 A - \sin^2 B + \sin^2 C \\ \ & = \frac{1- \cos 2A}{2} - \frac{1- \cos 2B}{2} + \sin^2 C\\ \ & = \frac{1- \cos 2A - (1- \cos 2B)}{2} + \sin^2 C\\ \ & = \frac{1- \cos 2A - 1+\cos 2B}{2} + \sin^2 C\\ \ & = \frac{ \cos 2B - \cos 2A}{2} + \sin^2 C\\ \ & = \frac{2\sin \frac{2A+2B}{2} \sin \frac{2A-2B}{2} }{2} + \sin^2 C\\ \ & = \frac{2\sin (A+B) \sin (A-B) }{2} + \sin^2 C \\ \ & = \sin (A+B) \sin (A-B) + \sin^2 C \\ \ & = \sin C \sin (A-B) + \sin^2 C\\ \ & = \sin C [ \sin (A-B) + \sin C ]\\ \ & = \sin C [ \sin (A - B) + \sin (A+B) ]\\ \ & = \sin C [ \sin (A + B) + \sin (A - B) ]\\ \ & = \sin C [ 2\sin A \cos B ]\\ \ & = 2\sin A \cos B \sin C \\ \ & = \text{RHS}\\ \end{align}
7. (a) If \( \alpha + \beta + \gamma = 180^{\circ}\), prove that: \( \sin^2\frac{\alpha}{2}-\sin^2\frac{\beta}{2}+\sin^2\frac{\gamma}{2} = 1 - 2\cos \frac{\alpha}{2} \sin \frac{\beta}{2} \cos \frac{\gamma}{2}\)
\begin{align} \text{Given, } \ & \\ \ & \alpha + \beta + \gamma = 180^{\circ}\\ \text{or, } \ & \frac { \alpha + \beta + \gamma}{2} = \frac{180^{\circ}}{2} \\ \text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ}\\ \text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} = 90^{\circ} - \frac{\gamma}{2} \\ \text{or, } \ & \sin \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \sin\left( 90^{\circ} - \frac{\gamma}{2} \right)=\cos \frac{ \gamma }{2} \\ \text{or, } \ & \cos \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \cos \left(90^{\circ} - \frac{\gamma}{2}\right)= \sin \frac{ \gamma }{2} \\ \text{Now} \ & \\ \text{LHS} \ & = \sin^2\frac{\alpha}{2}-\sin^2\frac{\beta}{2}+\sin^2\frac{\gamma}{2} \\ \ & = \frac{1- \cos \alpha }{2} - \frac{1- \cos \beta }{2} + \sin^2\frac{\gamma}{2} \\ \ & = \frac{1- \cos \alpha - (1- \cos \beta )}{2} + \sin^2\frac{\gamma}{2} \\ \ & = \frac{1- \cos \alpha - 1+ \cos \beta )}{2} + \sin^2\frac{\gamma}{2} \\ \ & = \frac{ \cos \beta - \cos \alpha}{2} + \sin^2\frac{\gamma}{2} \\ \ & = \frac{2\sin \frac{\alpha+\beta }{2} \sin \frac{\alpha -\beta}{2} }{2} + \sin^2\frac{\gamma}{2} \\ \ & = \frac{2\sin \left(\frac{\alpha}{2}+\frac{\beta}{2} \right) \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) }{2} + \sin^2\frac{\gamma}{2} \\ \ & = \sin \left(\frac{\alpha}{2}+\frac{\beta}{2} \right) \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) + 1 - \cos^2\frac{\gamma}{2} \\ \ & = 1+ \cos \frac{\gamma}{2} \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \cos^2\frac{\gamma}{2} \\ \ & = 1+ \cos \frac{\gamma}{2} \left[ \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \cos \frac{\gamma}{2} \right] \\ \ & = 1+ \cos \frac{\gamma}{2} \left[ \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \sin \left(\frac{\alpha}{2} + \frac{\beta}{2} \right) \right] \\ \ & = 1- \cos \frac{\gamma}{2} \left[ \sin \left(\frac{\alpha}{2} + \frac{\beta}{2} \right) - \sin \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) \right] \\ \ & = 1- \cos \frac{\gamma}{2} \left[2 \cos \frac{\alpha}{2} \sin \frac{\beta}{2} \right]\\ \ & = 1 - 2\cos \frac{\alpha}{2} \sin \frac{\beta}{2} \cos \frac{\gamma}{2}\\ \ & = \text{RHS}\\ \end{align}
7. (b) If \( \alpha + \beta + \gamma = 180^{\circ}\), prove that: \( \sin^2\frac{\alpha}{2}+\sin^2\frac{\beta}{2}-\sin^2\frac{\gamma}{2} = 1 - 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \sin \frac{\gamma}{2}\)
\begin{align} \text{Given, } \ & \\ \ & \alpha + \beta + \gamma = 180^{\circ}\\ \text{or, } \ & \frac { \alpha + \beta + \gamma}{2} = \frac{180^{\circ}}{2} \\ \text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ}\\ \text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} = 90^{\circ} - \frac{\gamma}{2} \\ \text{or, } \ & \sin \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \sin\left( 90^{\circ} - \frac{\gamma}{2} \right)=\cos \frac{ \gamma }{2} \\ \text{or, } \ & \cos \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \cos \left(90^{\circ} - \frac{\gamma}{2}\right)= \sin \frac{ \gamma }{2} \\ \text{Now} \ & \\ \text{LHS} \ & = \sin^2\frac{\alpha}{2}+\sin^2\frac{\beta}{2} -\sin^2\frac{\gamma}{2} \\ \ & = \frac{1- \cos \alpha }{2} + \frac{1- \cos \beta }{2} - \sin^2\frac{\gamma}{2} \\ \ & = \frac{1- \cos \alpha + 1 - \cos \beta }{2} - \sin^2\frac{\gamma}{2} \\ \ & = \frac{ 2 - \cos \beta - \cos \alpha}{2} - \sin^2\frac{\gamma}{2} \\ \ & = \frac{ 2 - (\cos \beta + \cos \alpha)}{2} - \sin^2\frac{\gamma}{2} \\ \ & = \frac{2-2\cos \frac{\alpha+\beta }{2} \cos \frac{\alpha -\beta}{2} }{2} - \sin^2\frac{\gamma}{2} \\ \ & = \frac{2-2\cos \left(\frac{\alpha}{2}+\frac{\beta}{2} \right) \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) }{2} - \sin^2\frac{\gamma}{2} \\ \ & = 1 -\cos \left(\frac{\alpha}{2}+\frac{\beta}{2} \right) \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \sin^2\frac{\gamma}{2} \\ \ & = 1 -\sin \frac{\gamma}{2} \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) - \sin^2\frac{\gamma}{2} \\ \ & = 1- \sin \frac{\gamma}{2} \left[ \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) + \sin \frac{\gamma}{2} \right] \\ \ & = 1 - \sin \frac{\gamma}{2} \left[ \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) + \cos \left(\frac{\alpha}{2} + \frac{\beta}{2} \right) \right] \\ \ & = 1- \sin \frac{\gamma}{2} \left[2 \cos \frac{\alpha}{2} \cos \frac{\beta}{2} \right]\\ \ & = 1 - 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \sin \frac{\gamma}{2}\\ \ & = \text{RHS}\\ \end{align}
7. (c) If \( \alpha + \beta + \gamma = 180^{\circ}\), prove that: \( -\cos^2\frac{\alpha}{2}+\cos^2\frac{\beta}{2}+\cos^2\frac{\gamma}{2} = 2\sin \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2}\)
\begin{align} \text{Given, } \ & \\ \ & \alpha + \beta + \gamma = 180^{\circ}\\ \text{or, } \ & \frac { \alpha + \beta + \gamma}{2} = \frac{180^{\circ}}{2} \\ \text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ}\\ \text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} = 90^{\circ} - \frac{\gamma}{2} \\ \text{or, } \ & \sin \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \sin\left( 90^{\circ} - \frac{\gamma}{2} \right)=\cos \frac{ \gamma }{2} \\ \text{or, } \ & \cos \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \cos \left(90^{\circ} - \frac{\gamma}{2}\right)= \sin \frac{ \gamma }{2} \\ \text{Now} \ & \\ \text{LHS} \ & = -\cos^2\frac{\alpha}{2}+\cos^2\frac{\beta}{2} + \cos^2\frac{\gamma}{2} \\ \ & = -\frac{1+ \cos \alpha }{2} + \frac{1+ \cos \beta }{2} + \cos^2\frac{\gamma}{2} \\ \ & = \frac{- (1+ \cos \alpha) + 1 + \cos \beta }{2} + \cos^2\frac{\gamma}{2} \\ \ & = \frac{ - 1 - \cos \alpha + 1 + \cos \beta }{2} + \cos^2\frac{\gamma}{2} \\ \ & = \frac{ \cos \beta - \cos \alpha}{2} + \cos^2\frac{\gamma}{2} \\ \ & = \frac{2\sin \frac{\alpha+\beta }{2} \sin \frac{\alpha -\beta}{2} }{2} + \cos^2\frac{\gamma}{2} \\ \ & = \sin \frac{\alpha+\beta }{2} \sin \frac{\alpha -\beta}{2} + \cos^2\frac{\gamma}{2} \\ \ & = \sin \left( \frac{\alpha}{2} + \frac{\beta}{2}\right) \sin \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\cos^2\frac{\gamma}{2} \\ \ & = \cos \frac{\gamma}{2} \sin \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\cos^2\frac{\gamma}{2} \\ \ & = \cos \frac{\gamma}{2} \left[ \sin \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\cos \frac{\gamma}{2} \right]\\ \ & = \cos \frac{\gamma}{2} \left[ \sin \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\sin \left( \frac{\alpha}{2} + \frac{\beta}{2}\right) \right] \\ \ & = \cos \frac{\gamma}{2} \left[ 2\sin \frac{\alpha}{2} \cos \frac{\beta}{2} \right]\\ \ & = 2\sin \frac{\alpha}{2} \cos \frac{\beta}{2} \cos \frac{\gamma}{2} \\ \ & = \text{RHS}\\ \end{align}
7. (d) If \( \alpha + \beta + \gamma = 180^{\circ}\), prove that: \( \cos^2\frac{\alpha}{2}+\cos^2\frac{\beta}{2}-\cos^2\frac{\gamma}{2} = 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \sin \frac{\gamma}{2}\)
\begin{align} \text{Given, } \ & \\ \ & \alpha + \beta + \gamma = 180^{\circ}\\ \text{or, } \ & \frac { \alpha + \beta + \gamma}{2} = \frac{180^{\circ}}{2} \\ \text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = 90^{\circ}\\ \text{or, } \ & \frac{\alpha}{2} + \frac{\beta}{2} = 90^{\circ} - \frac{\gamma}{2} \\ \text{or, } \ & \sin \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \sin\left( 90^{\circ} - \frac{\gamma}{2} \right)=\cos \frac{ \gamma }{2} \\ \text{or, } \ & \cos \left(\frac{\alpha}{2} + \frac{\beta}{2}\right) = \cos \left(90^{\circ} - \frac{\gamma}{2}\right)= \sin \frac{ \gamma }{2} \\ \text{Now} \ & \\ \text{LHS} \ & = \cos^2\frac{\alpha}{2}+\cos^2\frac{\beta}{2}-\cos^2\frac{\gamma}{2} \\ \ & = \frac{1+ \cos \alpha }{2} + \frac{1+ \cos \beta }{2} - \cos^2\frac{\gamma}{2} \\ \ & = \frac{ 1 + \cos \alpha + 1 + \cos \beta }{2} - \cos^2\frac{\gamma}{2} \\ \ & = \frac{ 2+ \cos \alpha + \cos \beta}{2} - \cos^2\frac{\gamma}{2} \\ \ & = \frac{2+ 2\cos \frac{\alpha+\beta }{2} \cos \frac{\alpha -\beta}{2} }{2} - \cos^2\frac{\gamma}{2} \\ \ & = 1+ \cos \frac{\alpha+\beta }{2} \cos \frac{\alpha -\beta}{2} - \cos^2\frac{\gamma}{2} \\ \ & = 1+ \cos \left( \frac{\alpha}{2} + \frac{\beta}{2}\right) \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right) - \left(1-\sin^2\frac{\gamma}{2}\right) \\ \ & = 1+ \sin \frac{\gamma}{2} \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right) - 1 + \sin^2\frac{\gamma}{2} \\ \ & = \sin \frac{\gamma}{2} \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right) + \sin^2\frac{\gamma}{2} \\ \ & = \sin \frac{\gamma}{2} \left[ \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\sin \frac{\gamma}{2} \right]\\ \ & = \sin \frac{\gamma}{2} \left[ \cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right)+\cos \left( \frac{\alpha}{2} + \frac{\beta}{2}\right) \right] \\ \ & = \sin \frac{\gamma}{2} \left[ \cos \left( \frac{\alpha}{2} + \frac{\beta}{2}\right)+\cos \left( \frac{\alpha}{2} - \frac{\beta}{2}\right) \right] \\ \ & = \sin \frac{\gamma}{2} \left[ 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \right]\\ \ & = 2\cos \frac{\alpha}{2} \cos \frac{\beta}{2} \sin \frac{\gamma}{2} \\ \ & = \text{RHS}\\ \end{align}
8. (a) If \( A+B+C = \pi^c\), prove that: \(\frac{\sin 2A + \sin 2B + \sin 2C}{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\)
8. (b) If \( A+B+C = \pi^c\), prove that: \(\frac{\sin 2A + \sin 2B + \sin 2C}{\sin A +\sin B + \sin C } = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}\)
\begin{align} \text{Given, } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \sin (A+ B) = \sin( \pi^c - C )=\sin C\\ \text{or, } \ & \cos (A+B) = \cos (\pi^c - C )=- \cos C\\ \text{Now} \ & \\ \text{LHS} \ & = \frac{\sin 2A + \sin 2B + \sin 2C}{\sin A +\sin B + \sin C } \\ \ & = \frac{2 \sin \frac{2A + 2B }{2} \cos \frac{2A-2B}{2}+ \sin 2C}{2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + \sin C } \\ \ & = \frac{2 \sin (A + B) \cos (A- B)+ \sin 2C}{ 2\sin \left( \frac{A}{2}+\frac{B}{2} \right)\cos \left( \frac{A}{2}-\frac{B}{2} \right) +\sin C } \\ \ & = \frac{2 \sin C \cos (A- B)+ 2\sin C \cos C}{2\cos \frac{C}{2} \cos \left( \frac{A}{2}-\frac{B}{2} \right) + 2\sin \frac{C}{2} \cos \frac{C}{2} } \\ \ & = \frac{2 \sin C \left[ \cos (A- B)+ \cos C \right]}{2\cos \frac{C}{2} \cos \left( \frac{A}{2}-\frac{B}{2} \right) + 2\sin \frac{C}{2} \cos \frac{C}{2} } \\ \ & = \frac{2 \sin C \left[ \cos (A- B) - \cos (A+B) \right]}{2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2}-\frac{B}{2} \right) + \sin \frac{C}{2} \right]} \\ \ & = \frac{2 \sin C \left[ 2\sin A \sin B \right]}{2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2}-\frac{B}{2} \right) + \cos \left( \frac{A}{2}+ \frac{B}{2} \right) \right] } \\ \ & = \frac{2 \sin C \left[ 2\sin A \sin B \right]}{2\cos \frac{C}{2} \left[ 2\cos \frac{A}{2} \cos \frac{C}{2} \right] } \\ \ & = \frac{4 \sin A \sin B \sin C }{4\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\ \ & = \frac{ \sin A \sin B \sin C }{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\ \ & = \frac{\left(2 \sin \frac{A}{2} \cos \frac{A}{2} \right) \left(2\sin \frac{B}{2} \cos \frac{B}{2} \right) \left( 2\sin \frac{C}{2} \cos \frac{C}{2} \right) }{\cos \frac{A}{2}\cos \frac{B}{2}\cos \frac{C}{2} } \\ \ & = 8\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} \\ \ & = \text{RHS}\\ \end{align}
8. (c) If \( A+B+C = \pi^c\), prove that: \(\frac{\sin A + \sin B - \sin C}{\sin A \sin B } = \sec \frac{A}{2} \sec \frac{B}{2} \cos \frac{C}{2} \)
8. (d) If \( A+B+C = \pi^c\), prove that: \(\frac{\sin^ 2A + \sin^ 2B - \sin^ 2C}{\sin A \sin B \sin C } = 2\cot C\)
9. (a) If \( X+Y+Z=180^{\circ}\), prove that: \( \sin X \cos Y \cos Z + \sin Y \cos Z \cos X + \sin Z cos X \cos Y = \sin X \sin Y \sin Z\)
9. (b) If \( X+Y+Z=180^{\circ}\), prove that: \( \cos X \sin Y \sin Z + \cos Y \sin Z \sin X + \cos Z \sin X \sin Y - \cos X \cos Y \cos Z =1\)
10. (a) If \( A+B+C = \pi^c\), prove that: \(\sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} = 1 + 4\sin \left(\frac{A+B}{4} \right) \sin \left(\frac{B+C}{4} \right) \sin \left(\frac{C+A}{4} \right) \)
\begin{align} \text{Given}\\ \ & A+B+C = \pi^c \\ \ & A + B = \pi^c - C \cdots (i) \\ \ & B + C = \pi^c - A \cdots (ii) \\ \ & C + A = \pi^ c - B \cdots (iii) \\ \text{LHS} \ & = \sin \frac{A}{2}+\sin \frac{B}{2}+\sin \frac{C}{2} \\ \ & = 2\sin \frac{\frac{A}{2} + \frac{B}{2}}{2} \cos \frac{\frac{A}{2} - \frac{B}{2}}{2} +\cos \left( \frac{\pi^c}{2} - \frac{C}{2} \right) \\ \ & = 2\sin \frac{A+B}{4} \cos \frac{A-B}{4} +\cos \frac{\pi^c-C}{2} \\ \ & = 2\sin \frac{\pi^c - C}{4} \cos \frac{A-B}{4} + 1-2\sin^2 \frac{\pi^c-C}{4}\\ \ & =1+ 2\sin \frac{\pi^c - C}{4} \cos \frac{A-B}{4} -2\sin^2 \frac{\pi^c-C}{4}\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ \cos \frac{A-B}{4} - \sin \frac{\pi^c+C}{4} \right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ \cos \frac{A-B}{4} - \cos \left( \frac{\pi^c}{2} - \frac{\pi^c-C}{4} \right) \right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ \cos \frac{A-B}{4} - \cos \left( \frac{2\pi^c-(\pi^c - C) }{4} \right) \right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ \cos \frac{A-B}{4} - \cos \frac{\pi^c + C }{4} \right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2 \sin \frac{ \frac{A-B}{4} + \frac{\pi^c + C }{4} }{2} \sin \frac{ \frac{\pi^c + C}{4} - \frac{A-B}{4} }{2} \right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ A-B+\pi^c + C }{4\times 2 } \sin \frac{\pi^c+C-A+B }{4\times 2}\right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ (A+C)+ \pi^c - B }{ 8 } \sin \frac{\pi^c -A+(B+C) }{ 8 }\right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c - B+ \pi^c - B }{ 8 } \sin \frac{\pi^c -A+\pi^c -A }{ 8 }\right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ 2\pi^c - 2B }{ 8 } \sin \frac{2\pi^c -2A}{ 8 }\right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ 2(\pi^c - B ) }{ 8 } \sin \frac{2( \pi^c -A )}{ 8 }\right]\\ \ & = 1+2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c - B }{ 4 } \sin \frac{ \pi^c -A }{ 4 }\right]\\ \ & = 1+ 4 \sin \frac{\pi^c -C}{4} \sin \frac{ \pi^c - B }{ 4 } \sin \frac{ \pi^c -A }{ 4 } \\ \ & = 1+ 4 \sin \frac{A+B}{4} \sin \frac{B+C }{ 4 } \sin \frac{ C+A}{ 4 } \\ \ & = \text{RHS}\\ \end{align}
10. (b) If \( A+B+C = \pi^c\), prove that:\( \cos \frac{A}{2} - \cos \frac{B}{2} + \cos \frac{C}{2} = 4\cos \frac{\pi^c +A}{4}\cos \frac{\pi^c -B}{4}\cos \frac{\pi^c +C}{4}\)
\begin{align} \text{Given}\\ \ & A+B+C = \pi^c \\ \ & A + B = \pi^c - C \cdots (i) \\ \ & B + C = \pi^c - A \cdots (ii) \\ \ & C + A = \pi^ c - B \cdots (iii) \\ \text{LHS} \ & = \cos \frac{A}{2} - \cos \frac{B}{2} + \cos \frac{C}{2} \\ \ & = 2\sin \frac{\frac{A}{2} + \frac{B}{2}}{2} \sin \frac{\frac{B}{2} - \frac{A}{2}}{2} +\sin \left( \frac{\pi^c}{2} - \frac{C}{2} \right) \\ \ & = 2\sin \frac{A+B}{4} \sin \frac{B-A}{4} +\sin \frac{\pi^c-C}{2} \\ \ & = 2\sin \frac{A+B}{4} \sin \frac{B-A}{4} +\sin 2\left(\frac{\pi^c-C}{4} \right) \\ \ & = 2\sin \frac{\pi^c - C}{4} \sin \frac{B-A}{4} + 2\sin \frac{\pi^c-C}{4}\cos \frac{\pi^c-C}{4} \\ \ & = 2\sin \frac{\pi^c -C}{4} \left[ \sin \frac{B-A}{4} + \cos \frac{\pi^c - C}{4} \right]\\ \ & = 2\sin \frac{\pi^c -C}{4} \left[ \sin \frac{B-A}{4} + \sin \left( \frac{\pi^c}{2} - \frac{\pi^c-C}{4} \right) \right]\\ \ & =2\sin \frac{\pi^c -C}{4} \left[ \sin \frac{B-A}{4} + \sin \left( \frac{2\pi^c-(\pi^c - C) }{4} \right) \right]\\ \ & = 2\sin \frac{\pi^c -C}{4} \left[ \sin \frac{B-A}{4} + \sin \frac{\pi^c + C }{4} \right]\\ \ & = 2\sin \frac{\pi^c -C}{4} \left[ 2 \sin \frac{ \frac{B-A}{4} + \frac{\pi^c + C }{4} }{2} \cos \frac{ \frac{\pi^c + C}{4} - \frac{B-A}{4} }{2} \right]\\ \ & = 2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ B-A+\pi^c + C }{4\times 2 } \cos \frac{\pi^c+C - B+A }{4\times 2}\right]\\ \ & = 2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c -A + ( B+C) }{ 8 } \cos \frac{\pi^c -B+(A+C) }{ 8 }\right]\\ \ & =2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c - A+ \pi^c - A }{ 8 } \cos \frac{\pi^c -B+\pi^c -B }{ 8 }\right]\\ \ & = 2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ 2\pi^c - 2A }{ 8 } \cos \frac{2\pi^c -2B}{ 8 }\right]\\ \ & =2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ 2(\pi^c - A ) }{ 8 } \cos \frac{2( \pi^c -B )}{ 8 }\right]\\ \ & =2\sin \frac{\pi^c -C}{4} \left[ 2\sin \frac{ \pi^c - A }{ 4 } \cos \frac{ \pi^c -B }{ 4 }\right]\\ \ & = 4 \sin \frac{\pi^c -A}{4} \cos \frac{ \pi^c - B }{ 4 } \sin \frac{ \pi^c -C }{ 4 } \\ \ & = 4 \cos\left( \frac{\pi^c}{2} - \frac{\pi^c -A}{4} \right) \cos \frac{ \pi^c - B }{ 4 } \cos \left( \frac{ \pi^c }{2} - \frac{ \pi^c -C }{ 4 }\right) \\ \ & = 4 \cos \frac{ 2\pi^c - \pi^c + A}{4} \cos \frac{\pi^c - B}{4} \cos \frac{ 2\pi^c - \pi^c + C } { 4} \\ \ & = 4 \cos \frac{ \pi^c + A}{4} \cos \frac{\pi^c - B}{4} \cos \frac{ \pi^c + C } { 4} \\ \ & = \text{RHS}\\ \end{align}
10. (c) If \( A+B+C = \pi^c\), prove that: \(\sin A + \sin B +\sin C = 4\sin \frac{B+C}{2}\sin \frac{C+A}{2}\sin \frac{A+B}{2}\)
\begin{align} \text{Given } \ & \\ \ & A+B+C = \pi^c \\ \text{or, } \ & \frac{A+B+C }{2} =\frac{\pi^c}{2}\\ \text{or, } \ & \frac{A}{2} + \frac{B}{2} = \frac{\pi^c}{2}- \frac{C}{2}\\ \text{or, } \ & \sin\left(\frac{A}{2} + \frac{B}{2} \right) = \sin \left(\frac{\pi^c}{2}- \frac{C}{2} \right) = \cos \frac{C}{2}\\ \text{or, } \ & \cos\left(\frac{A}{2} + \frac{B}{2} \right) = \cos \left(\frac{\pi^c}{2}- \frac{C}{2} \right) = \sin \frac{C}{2}\\ \text{Now} \ & \\ \text{LHS} \ & = \sin A + \sin B + \sin C\\ \ & = 2\sin \frac{A+B}{2} \cos \frac{A-B}{2} + \sin C\\ \ & = 2 \sin \left( \frac{A}{2} + \frac{B}{2} \right) \cos \left( \frac{A}{2} + \frac{B}{2} \right) + 2\sin \frac{C}{2}\cos \frac{C}{2} \\ \ & = 2\cos \frac{C}{2} \cos \left( \frac{A}{2} - \frac{B}{2} \right) + 2\sin \frac{C}{2}\cos \frac{C}{2} \\ \ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \sin \frac{C}{2} \right] \\ \ & = 2\cos \frac{C}{2} \left[ \cos \left( \frac{A}{2} - \frac{B}{2} \right) + \cos \left( \frac{A}{2} + \frac{B}{2} \right) \right] \\ \ & = 2\cos \frac{C}{2} \left[ 2\cos \frac{A}{2} \cos \frac{B}{2}\right] \\ \ & = 4\cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} \\ \ & = 4\sin \left( \frac{ \pi^c}{2} - \frac{A}{2} \right)\sin \left( \frac{ \pi^c}{2} - \frac{B}{2} \right)\sin \left( \frac{ \pi^c}{2} - \frac{C}{2} \right) \\ \ & = 4\sin \frac{\pi^c - A}{2} \sin \frac{\pi^c - B}{2} \sin \frac{\pi^c - C}{2} \\ \ & = 4\sin \frac{B+C}{2}\sin \frac{C+A}{2}\sin \frac{A+B}{2}[ \ \ \because A + B + C = \pi^c \ \ ] \\ \ & = \text{RHS}\\ \end{align}
10. (d) If \( A+B+C = \pi^c\), prove that: \( \cos A + \cos B + \cos C = 1+ 4\cos \frac{\pi^c -A }{2}\cos \frac{\pi^c - B}{2}\cos \frac{\pi^c - C}{2} \)
1 Comments
Good job sir
ReplyDelete