Question

If $\cos y\cos (\frac{\pi }{2}-x)-\cos (\frac{\pi }{2}-y)\cos x+\sin y\cos (\frac{\pi }{2}-x)+\cos x\sin (\frac{\pi }{2}-y)=0$, then which of the following is correct

• A. $x=n\pi +y,n\in Z$
• B. $x=y+\frac{\pi }{4},n\in Z$
• C. $x=y$
• D. $x=n\pi +y-\frac{\pi }{4},n\in Z$

Answer 1

The correct option is D $x=n\pi +y-\frac{\pi }{4},n\in Z$

$\cos y\cos (\frac{\pi }{2}-x)-\cos (\frac{\pi }{2}-y)\cos x+\sin y\cos (\frac{\pi }{2}-x)+\cos x\sin (\frac{\pi }{2}-y)=0\Rightarrow \cos y\sin x-\sin y\cos x+\sin y\sin x+\cos x\cos y\Rightarrow \sin (x-y)+\cos (x-y)=0\Rightarrow \tan (x-y)=-1=-\tan \frac{\pi }{4}\Rightarrow x-y=n\pi -\frac{\pi }{4}\Rightarrow x=n\pi +y-\frac{\pi }{4},n\in Z$