If tan (pi cos theta ) = cot (pi sin theta ), | vedclass

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(C) Given $	an (\pi \cos 	heta ) = \cot (\pi \sin 	heta )$.Using the identity $\cot(x) = 	an(\frac{\pi}{2} - x)$,we have $	an (\pi \cos 	heta )

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$\frac{1}{2\sqrt{2}}$

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(C) Given $	an (\pi \cos 	heta ) = \cot (\pi \sin 	heta )$.Using the identity $\cot(x) = 	an(\frac{\pi}{2} - x)$,we have $	an (\pi \cos 	heta ) = 	an \left( \frac{\pi}{2} - \pi \sin 	heta ight)$.This implies $\pi \cos 	heta = n\pi + \frac{\pi}{2} - \pi \sin 	heta$ for some integer $n$.Dividing by $\pi$,we get $\cos 	heta + \sin 	heta = n + \frac{1}{2}$.Since the range of $\sin 	heta + \cos 	heta$ is $[-\sqrt{2}, \sqrt{2}] \approx [-1.414, 1.414]$,the only possible value for $n$ is $0$,giving $\sin 	heta + \cos 	heta = \frac{1}{2}$.We need to find $\sin \left( 	heta + \frac{\pi}{4} ight) = \sin 	heta \cos \frac{\pi}{4} + \cos 	heta \sin \frac{\pi}{4}$.$= \frac{1}{\sqrt{2}} (\sin 	heta + \cos 	heta) = \frac{1}{\sqrt{2}} 	imes \frac{1}{2} = \frac{1}{2\sqrt{2}}$.

If $\tan (\pi \cos \theta ) = \cot (\pi \sin \theta )$,then $\sin \left( \theta + \frac{\pi }{4} \right)$ equals

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