A) \[\frac{1}{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) \[\pm \frac{1}{2\sqrt{2}}\]
D) None of these
Correct Answer: C
Solution :
We have,\[\tan (\pi \cos \theta )=\cot (\pi \sin \theta )\] \[\Rightarrow \] \[\tan (\pi \cos \theta )=\tan \left( \frac{\pi }{2}-\pi \sin \theta \right)\] \[\Rightarrow \] \[\pi \cos \theta =\frac{\pi }{2}-\pi \sin \theta +n\pi ;\,\,n\in Z\] \[\Rightarrow \] \[\cos \theta +\sin \theta =\frac{1}{2}+n,\,\,n\in Z\] \[\Rightarrow \] \[\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{\sqrt{2}}\sin \theta =\frac{2n+1}{2\sqrt{2}},\,\,n\in Z\] \[\Rightarrow \] \[\cos \left( \theta -\frac{\pi }{4} \right)=\frac{2n+1}{2\sqrt{2}},\,\,n\in Z\] \[\Rightarrow \] \[\cos \left( \theta -\frac{\pi }{4} \right)=\pm \frac{1}{2\sqrt{2}}\] [for\[n=0\]and\[n=-1]\]
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