If tan(A + B) = p and tan(A − B) = q, then find tan 2A If \( \tan(A+B)=p \) and \( \tan(A-B)=q \), then find \( \tan2A \) Solution: Since, \[ 2A=(A+B)+(A-B) \] Using identity, \[ \tan(x+y) = \frac{\tan x+\tan y}{1-\tan x\tan y} \] \[ \tan2A = \tan[(A+B)+(A-B)] \] \[ = \frac{\tan(A+B)+\tan(A-B)} {1-\tan(A+B)\tan(A-B)} \] […]

The value of (sin 70° + cos 40°)/(cos 70° + sin 40°) The value of \( \dfrac{\sin70^\circ+\cos40^\circ}{\cos70^\circ+\sin40^\circ} \) is Solution: Using, \[ \cos\theta=\sin(90^\circ-\theta) \] \[ \cos40^\circ=\sin50^\circ \] and \[ \cos70^\circ=\sin20^\circ \] Therefore, \[ = \frac{\sin70^\circ+\sin50^\circ} {\sin20^\circ+\sin40^\circ} \] Using identity, \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ = \frac{ 2\sin60^\circ\cos10^\circ } { 2\sin30^\circ\cos10^\circ } \]

The value of sin(π/18) + sin(π/9) + sin(2π/9) + sin(5π/18) The value of \( \sin\frac{\pi}{18}+\sin\frac{\pi}{9}+\sin\frac{2\pi}{9}+\sin\frac{5\pi}{18} \) is given by Options: (a) \( \sin\frac{7\pi}{18}+\sin\frac{4\pi}{9} \) (b) \(1\) (c) \( \cos\frac{\pi}{6}+\cos\frac{3\pi}{7} \) (d) \( \cos\frac{\pi}{9}+\sin\frac{\pi}{9} \) Solution: \[ =\sin\frac{\pi}{18}+\sin\frac{5\pi}{18} + \sin\frac{\pi}{9}+\sin\frac{2\pi}{9} \] Using identity, \[ \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ = 2\sin\frac{6\pi}{36}\cos\frac{4\pi}{36} + 2\sin\frac{3\pi}{18}\cos\frac{\pi}{18} \]

If sin x + sin y = √3(cos y − cos x), then find sin 3x + sin 3y If \( \sin x+\sin y=\sqrt3(\cos y-\cos x) \), then \( \sin3x+\sin3y \) is Options: (a) \(2\sin3x\) (b) \(0\) (c) \(1\) (d) none of these Solution: Using identities, \[ \sin x+\sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2} \] and \[

If sin(B + C − A), sin(C + A − B), sin(A + B − C) are in A.P., then cot A, cot B, cot C are in If \( \sin(B+C-A), \sin(C+A-B), \sin(A+B-C) \) are in A.P., then \( \cot A, \cot B, \cot C \) are in Options: (a) GP (b) HP (c) AP

If A, B, C are in A.P., then find (sin A − sin C)/(cos C − cos A) If \( A, B, C \) are in A.P., then \( \dfrac{\sin A-\sin C}{\cos C-\cos A} \) is Options: (a) \( \tan B \) (b) \( \cot B \) (c) \( \tan2B \) (d) none of these

If cos A = m cos B, then find cot((A + B)/2) cot((B − A)/2) If \( \cos A = m\cos B \), then \( \cot\frac{A+B}{2}\cot\frac{B-A}{2} \) is Options: (a) \( \frac{m-1}{m+1} \) (b) \( \frac{m+2}{m-2} \) (c) \( \frac{m+1}{m-1} \) (d) none of these Solution: Given, \[ \cos A=m\cos B \] \[ \frac{\cos A}{\cos

sin 47° + sin 61° − sin 11° − sin 25° \( \sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ \) is equal to Options: (a) \( \sin36^\circ \) (b) \( \cos36^\circ \) (c) \( \sin7^\circ \) (d) \( \cos7^\circ \) Solution: \[ =(\sin47^\circ-\sin11^\circ)+(\sin61^\circ-\sin25^\circ) \] Using identity, \[ \sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \] \[ = 2\cos29^\circ\sin18^\circ + 2\cos43^\circ\sin18^\circ \] \[ =

The value of sin 50° − sin 70° + sin 10° is equal to The value of \( \sin50^\circ-\sin70^\circ+\sin10^\circ \) is equal to Options: (a) \(1\) (b) \(0\) (c) \( \frac12 \) (d) \(2\) Solution: \[ =\sin50^\circ-\sin70^\circ+\sin10^\circ \] Grouping first two terms, \[ =(\sin50^\circ-\sin70^\circ)+\sin10^\circ \] Using identity, \[ \sin A-\sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} \] \[

cos 35° + cos 85° + cos 155° \( \cos35^\circ+\cos85^\circ+\cos155^\circ \) Options: (a) \(0\) (b) \( \frac1{\sqrt3} \) (c) \( \frac1{\sqrt2} \) (d) \( \cos275^\circ \) Solution: \[ =\cos35^\circ+\cos85^\circ+\cos155^\circ \] Using, \[ \cos(180^\circ-\theta)=-\cos\theta \] \[ =\cos35^\circ+\cos85^\circ-\cos25^\circ \] Using identity, \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ = 2\cos60^\circ\cos25^\circ-\cos25^\circ \] \[ = 2\left(\frac12\right)\cos25^\circ-\cos25^\circ \] \[