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

Jharkhand CECEJharkhand CECE 2010

Solution:

We have, $ \cos y\cos \left( \frac{\pi }{2}-x \right)-\cos \left( \frac{\pi }{2}-y \right)\cos x $
$ +\sin y\cos \left( \frac{\pi }{2}-x \right)+\cos x\sin \left( \frac{\pi }{2}-y \right)=0 $
$ \Rightarrow $ $ \cos y\sin x-\sin y\cos x+\sin y\sin x $
$ +\cos x\cos y=0 $
$ \Rightarrow $ $ (\sin x\cos y-\cos x\sin y)+(\cos x\cos y $
$ +\sin x\sin y)=0 $
$ \Rightarrow $ $ \sin (x-y)+\cos (x-y)=0 $
$ \Rightarrow $ $ \sin (x-y)=-\cos (x-y) $
$ \Rightarrow $ $ \tan (x-y)=-1 $
$ \Rightarrow $ $ x-y=n\pi -\frac{\pi }{4}\Rightarrow x=n\pi -\frac{\pi }{4}+y $