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. $x=n\pi -\frac{\pi }{4}+y$

Answer 1

Answer: (D)

$x=n\pi -\frac{\pi }{4}+y$

$\cos y\cos (\frac{\pi }{2}-x)-\cos (\frac{\pi }{2}-y)\cos x+\sin y\cos (\frac{\pi }{2}-x)+\cos x\sin (\frac{\pi }{2}-y)=0\Rightarrow \cos y\sin x-\sin y\cos x+\sin y\sin x+\cos x\cos y=0\Rightarrow \sin (x-y)+\cos (x-y)=0\Rightarrow \sin (x-y)-\cos (x-y)\Rightarrow \frac{\sin (x-y)}{\cos (x-y)}=-1\Rightarrow \tan (x-y)=\tan (\pi -\frac{\pi }{4})\Rightarrow (x-y)=\pi -\frac{\pi }{4};also(x-y)=2\pi -\frac{\pi }{4}andsoon\therefore (x-y)=n\pi -\frac{\pi }{4}\Rightarrow x=n\pi -\frac{\pi }{4}+y$