A) \[x=0\]
B) \[y=0\]
C) \[x=y\]
D) None of these
Correct Answer: D
Solution :
We have,\[\cos y\cos \left( \frac{\pi }{2}-x \right)-\cos \left( \frac{\pi }{2}-y \right)\cos x\] \[+\sin y\cos \left( \frac{\pi }{2}-x \right)+\cos x\sin \left( \frac{\pi }{2}-y \right)=0\] \[\Rightarrow \]\[\cos y\sin x-\sin y\cos x+\sin y\sin x\] \[+\cos x\cos y=0\] \[\Rightarrow \]\[(\sin x\cos y-\cos x\sin y)+(\cos x\cos y\] \[+\sin x\sin y)=0\] \[\Rightarrow \] \[\sin (x-y)+\cos (x-y)=0\] \[\Rightarrow \] \[\sin (x-y)=-\cos (x-y)\] \[\Rightarrow \] \[\tan (x-y)=-1\] \[\Rightarrow \] \[x-y=n\pi -\frac{\pi }{4}\Rightarrow x=n\pi -\frac{\pi }{4}+y\]
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