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

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

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$\frac{1}{\sqrt{7}}$

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(A) Given $	an(\pi \sin 	heta) = \cot(\pi \cos 	heta)$.Using $\cot(x) = 	an(\frac{\pi}{2} - x)$,we have $	an(\pi \sin 	heta) = 	an(\frac{\pi}{2} - \pi \cos 	heta)$.This implies $\pi \sin 	heta = n\pi + \frac{\pi}{2} - \pi \cos 	heta$ for some integer $n$.$\sin 	heta + \cos 	heta = n + \frac{1}{2}$.Since $- \sqrt{2} \le \sin 	heta + \cos 	heta \le \sqrt{2}$,the possible values for $n$ are $0$ or $-1$.If $n = 0$,$\sin 	heta + \cos 	heta = \frac{1}{2}$.$\sqrt{2} \sin(	heta + \frac{\pi}{4}) = \frac{1}{2} \Rightarrow \sin(	heta + \frac{\pi}{4}) = \frac{1}{2\sqrt{2}}$.Using $\cos^2 x = 1 - \sin^2 x$,$\cos^2(	heta + \frac{\pi}{4}) = 1 - \frac{1}{8} = \frac{7}{8}$.$\cot^2(	heta + \frac{\pi}{4}) = \frac{\cos^2(	heta + \frac{\pi}{4})}{\sin^2(	heta + \frac{\pi}{4})} = \frac{7/8}{1/8} = 7$.Since $\cot(	heta - \frac{\pi}{4}) = 	an(\frac{\pi}{2} - (	heta - \frac{\pi}{4})) = 	an(\frac{3\pi}{4} - 	heta)$,we use the identity $\cot(	heta - \frac{\pi}{4}) = -	an(	heta + \frac{\pi}{4})$.Thus,$\left| \cot(	heta - \frac{\pi}{4}) ight| = \left| 	an(	heta + \frac{\pi}{4}) ight| = \frac{1}{\sqrt{7}}$.

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

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