The value of cos y cos (frac{pi}{2} - x) - co | vedclass

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

Question

(D) Given expression: $\cos y \cos (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - y) \cos x + \sin y \cos (\frac{\pi}{2} - x) + \cos x \sin (\frac{\pi}{2

Vedclass · Accepted Answer

$x = n\pi - \frac{\pi}{4} + y, (n \in I)$

Vedclass · Answer

(D) Given expression: $\cos y \cos (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - y) \cos x + \sin y \cos (\frac{\pi}{2} - x) + \cos x \sin (\frac{\pi}{2} - y) = 0$Using $\cos (\frac{\pi}{2} - 	heta) = \sin 	heta$ and $\sin (\frac{\pi}{2} - 	heta) = \cos 	heta$,the expression becomes:$\cos y \sin x - \sin y \cos x + \sin y \sin x + \cos x \cos y = 0$Rearranging the terms:$(\sin x \cos y - \cos x \sin y) + (\sin x \sin y + \cos x \cos y) = 0$Using the identities $\sin(A - B) = \sin A \cos B - \cos A \sin B$ and $\cos(A - B) = \cos A \cos B + \sin A \sin B$:$\sin(x - y) + \cos(x - y) = 0$Dividing by $\cos(x - y)$ (assuming $\cos(x - y) 
eq 0$):$	an(x - y) = -1$$	an(x - y) = 	an(-\frac{\pi}{4})$$x - y = n\pi - \frac{\pi}{4}$$x = n\pi - \frac{\pi}{4} + y, (n \in I)$

The value of $\cos y \cos (\frac{\pi}{2} - x) - \cos (\frac{\pi}{2} - y) \cos x + \sin y \cos (\frac{\pi}{2} - x) + \cos x \sin (\frac{\pi}{2} - y)$ is zero,if

Similar Questions

The number of solutions of the given equation $\tan \theta + \sec \theta = \sqrt{3}$,where $0 < \theta < 2\pi$ is

If $\cos \theta = \frac{-1}{2}$ and $0^o < \theta < 360^o$,then the values of $\theta$ are

The general value of $\theta$ satisfying the equation $2\sin^2 \theta - 3\sin \theta - 2 = 0$ is

If $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$,then $\sin \left( \theta + \frac{\pi }{4} \right)$ equals

The sum of the solutions in $(0, 2\pi)$ for the equation $\cos x \cos \left(\frac{\pi}{3}-x\right) \cos \left(\frac{\pi}{3}+x\right) = \frac{1}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module