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

Prove that $\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x = 1$.

If $u = \log \tan \left(\frac{\pi}{4} + \frac{\theta}{2}\right)$,then $\cosh u =$

If $4 \cos x + 3 \sin x = 5$,then find the value of $\tan x$.

If $\cot x \cot y = a$ and $x+y = \frac{\pi}{6}$,then the quadratic equation whose roots are $\cot x$ and $\cot y$ is

If $2 \sin \theta + 3 \cos \theta = 2$ and $\theta \neq (2n + 1) \frac{\pi}{2}$,then find the value of $3 \sin \theta - 2 \cos \theta$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module