Question

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 :

• A. $x=0$
• B. $y=0$
• C. $x=y$
• D. $x=n\pi -\frac{\pi }{4}+y;\left(nϵI\right)$

Answer 1

The correct option is D $x=n\pi -\frac{\pi }{4}+y;\left(nϵI\right)$

Given, $\cos y\cos (\frac{\pi }{2}-x)-\cos x\cos (\frac{\pi }{2}-y)+\sin y\cos (\frac{\pi }{2}-x)+\cos x\sin (\frac{\pi }{2}-y)$
$\Rightarrow \cos y\sin x-\sin y\cos x+\sin x\sin y+\cos x\cos y=0\Rightarrow \sin (x-y)+\cos (x-y)=0$
$\Rightarrow \sin (\frac{\pi }{4}+(x-y))=0\Rightarrow \frac{\pi }{4}+(x-y)=n\pi \Rightarrow x=n\pi +y$