If sin x + sin y = p and cos x + cos y = q,th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: $\sin x + \sin y = p$ $(i)$ and $\cos x + \cos y = q$ (ii). Squaring and adding $(i)$ and (ii): $(\sin x + \sin y)^2 + (\cos x + \cos y)^2

Vedclass · Accepted Answer

$\frac{p^2 + q^2}{q^2 - p^2}$

Vedclass · Answer

(B) Given: $\sin x + \sin y = p$ $(i)$ and $\cos x + \cos y = q$ (ii). Squaring and adding $(i)$ and (ii): $(\sin x + \sin y)^2 + (\cos x + \cos y)^2 = p^2 + q^2$ $(\sin^2 x + \cos^2 x) + (\sin^2 y + \cos^2 y) + 2(\sin x \sin y + \cos x \cos y) = p^2 + q^2$ $1 + 1 + 2 \cos(x - y) = p^2 + q^2 \implies 2 + 2 \cos(x - y) = p^2 + q^2$ (iii). Now,consider $q^2 - p^2 = (\cos x + \cos y)^2 - (\sin x + \sin y)^2$ $= (\cos^2 x - \sin^2 x) + (\cos^2 y - \sin^2 y) + 2(\cos x \cos y - \sin x \sin y)$ $= \cos 2x + \cos 2y + 2 \cos(x + y)$ $= 2 \cos(x + y) \cos(x - y) + 2 \cos(x + y) = 2 \cos(x + y) [\cos(x - y) + 1]$. From (iii),$2 \cos(x - y) + 2 = p^2 + q^2$,so $\cos(x - y) + 1 = \frac{p^2 + q^2}{2}$. Substituting this into the expression for $q^2 - p^2$: $q^2 - p^2 = 2 \cos(x + y) \cdot \frac{p^2 + q^2}{2} = \cos(x + y)(p^2 + q^2)$. Therefore,$\sec(x + y) = \frac{1}{\cos(x + y)} = \frac{p^2 + q^2}{q^2 - p^2}$. Thus,option $B$ is correct.

If $\sin x + \sin y = p$ and $\cos x + \cos y = q$,then $\sec(x + y) = $

Similar Questions

$\cot 18^{\circ} \cdot \cot 36^{\circ}+1=$

If $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2-px+q=0$,then $\sin^2(A+B)$ is equal to

If $\tan \theta + \cot \theta = 4$,then $\tan^{4} \theta + \cot^{4} \theta = $

$3\left[ \sin^4\left( \frac{3\pi}{2} - \alpha \right) + \sin^4(3\pi + \alpha) \right] - 2\left[ \sin^6\left( \frac{\pi}{2} + \alpha \right) + \sin^6(5\pi - \alpha) \right] = $

$\sin \frac{2 \pi}{5}+\sin \frac{4 \pi}{5}+\sin \frac{6 \pi}{5}+\sin \frac{8 \pi}{5}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module