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

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(D) Given $	an (\pi \cos 	heta) = \cot (\pi \sin 	heta)$.We know that $\cot x = 	an \left(\frac{\pi}{2} - xight)$.So,$	an (\pi \cos 	heta) = \

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

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(D) Given $	an (\pi \cos 	heta) = \cot (\pi \sin 	heta)$.We know that $\cot x = 	an \left(\frac{\pi}{2} - xight)$.So,$	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 $\cos 	heta + \sin 	heta = n + \frac{1}{2}$.Multiplying by $\frac{1}{\sqrt{2}}$,we get $\frac{1}{\sqrt{2}} \cos 	heta + \frac{1}{\sqrt{2}} \sin 	heta = \frac{n + 0.5}{\sqrt{2}}$.$\sin \left(\frac{\pi}{4} + 	hetaight) = \frac{n + 0.5}{\sqrt{2}}$.For the sine function to be within $[-1, 1]$,we test $n=0$: $\sin \left(\frac{\pi}{4} + 	hetaight) = \frac{0.5}{\sqrt{2}} = \frac{1}{2 \sqrt{2}}$.

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

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