Ravi Kant Kumar

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, \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ = 2\cos60^\circ\cos25^\circ-\cos25^\circ \] \[ = 2\left(\frac12\right)\cos25^\circ-\cos25^\circ \] \[

If sin α + sin β = a and cos α − cos β = b, then find tan((α − β)/2) If \( \sin\alpha+\sin\beta=a \) and \( \cos\alpha-\cos\beta=b \), then \( \tan\frac{\alpha-\beta}{2} \) is Options: (a) \( -\frac{a}{b} \) (b) \( -\frac{b}{a} \) (c) \( \sqrt{a^2+b^2} \) (d) none of these Solution: Using identities, \[

The value of sin 78° − sin 66° − sin 42° + sin 6° is The value of \( \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ \) is Options: (a) \( \frac12 \) (b) \( -\frac12 \) (c) \( -1 \) (d) none of these Solution: \[ =\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ \] Grouping terms, \[ =(\sin78^\circ-\sin66^\circ)-(\sin42^\circ-\sin6^\circ) \] Using identity, \[ \sin A-\sin B =

The Value of cos 52° + cos 68° + cos 172° \( \cos52^\circ+\cos68^\circ+\cos172^\circ \) Options: (a) \(0\) (b) \(1\) (c) \(2\) (d) \( \frac32 \) Solution: Using, \[ \cos(180^\circ-\theta)=-\cos\theta \] \[ \cos172^\circ=-\cos8^\circ \] Therefore, \[ \cos52^\circ+\cos68^\circ+\cos172^\circ \] \[ =\cos52^\circ+\cos68^\circ-\cos8^\circ \] Using identity, \[ \cos A+\cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} \] \[ = 2\cos60^\circ\cos8^\circ-\cos8^\circ \] \[ =

If sin 2θ + sin 2ϕ = 1/2 and cos 2θ + cos 2ϕ = 3/2, then find cos²(θ − ϕ) If \( \sin2\theta+\sin2\phi=\frac12 \) and \( \cos2\theta+\cos2\phi=\frac32 \), then find \( \cos^2(\theta-\phi) \) Options: (a) \( \frac38 \) (b) \( \frac58 \) (c) \( \frac34 \) (d) \( \frac54 \) Solution: Using, \[ \sin

sin 163° cos 347° + sin 73° sin 167° \( \sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ \) Options: (a) \(0\) (b) \(\frac12\) (c) \(1\) (d) none of these Solution: \[ =\sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ \] \[ =\sin17^\circ\cos13^\circ+\sin73^\circ\sin13^\circ \] \[ =\sin17^\circ\cos13^\circ+\cos17^\circ\sin13^\circ \] Using identity, \[ \sin A\cos B+\cos A\sin B=\sin(A+B) \] \[ =\sin(17^\circ+13^\circ) \] \[ =\sin30^\circ \] \[ =\frac12 \] \[ \boxed{\frac12} \] Correct

cos 40° + cos 80° + cos 160° + cos 240° \( \cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ \) Options: (a) \(0\) (b) \(1\) (c) \(\frac12\) (d) \(-\frac12\) Solution: We know that, \[ \cos(180^\circ-\theta)=-\cos\theta \] and \[ \cos(180^\circ+\theta)=-\cos\theta \] Therefore, \[ \cos160^\circ=-\cos20^\circ \] and \[ \cos240^\circ=-\cos60^\circ=-\frac12 \] Hence, \[ \cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ \] \[ =\cos40^\circ+\cos80^\circ-\cos20^\circ-\frac12 \] Using identity, \[ \cos20^\circ=\cos40^\circ+\cos80^\circ \] Substituting,

Prove that tan x tan(π/3 − x) tan(π/3 + x) = tan 3x Prove that: \( \tan x \tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right)=\tan3x \) Solution: \[ \tan x \tan\left(\frac{\pi}{3}-x\right)\tan\left(\frac{\pi}{3}+x\right) \] Using, \[ \tan\left(\frac{\pi}{3}-x\right)=\frac{\sqrt3-\tan x}{1+\sqrt3\tan x} \] and \[ \tan\left(\frac{\pi}{3}+x\right)=\frac{\sqrt3+\tan x}{1-\sqrt3\tan x} \] \[ =\tan x\cdot \frac{\sqrt3-\tan x}{1+\sqrt3\tan x} \cdot \frac{\sqrt3+\tan x}{1-\sqrt3\tan x} \] \[ =\tan x\cdot \frac{3-\tan^2x}{1-3\tan^2x} \] \[

If α + β = π/2, Show that the Maximum Value of cos α cos β is 1/2 If \( \alpha+\beta=\frac{\pi}{2} \), show that the maximum value of \( \cos\alpha\cos\beta \) is \( \frac12 \) Solution: Given, \[ \alpha+\beta=\frac{\pi}{2} \] Using identity, \[ 2\cos\alpha\cos\beta=\cos(\alpha+\beta)+\cos(\alpha-\beta) \] \[ 2\cos\alpha\cos\beta = \cos\frac{\pi}{2}+\cos(\alpha-\beta) \] \[ 2\cos\alpha\cos\beta = 0+\cos(\alpha-\beta) \]

Show that sin(B − C) cos(A − D) + sin(C − A) cos(B − D) + sin(A − B) cos(C − D) = 0 Show that: \( \sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D)+\sin(A-B)\cos(C-D)=0 \) Solution: \[ \sin(B-C)\cos(A-D)+\sin(C-A)\cos(B-D) \] \[ +\sin(A-B)\cos(C-D) \] Using identity, \[ \sin x \cos y=\frac12[\sin(x+y)+\sin(x-y)] \] \[ =\frac12[\sin(B-C+A-D)+\sin(B-C-A+D)] \] \[ +\frac12[\sin(C-A+B-D)+\sin(C-A-B+D)] \] \[ +\frac12[\sin(A-B+C-D)+\sin(A-B-C+D)] \] \[ =\frac12[\sin(A+B-C-D)+\sin(B+D-A-C)