If sin left( frac{pi }{4} cot theta right) = | vedclass

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(A) Given $\sin \left( \frac{\pi }{4} \cot 	heta ight) = \cos \left( \frac{\pi }{4} 	an 	heta ight)$.Using the identity $\cos x = \sin \left( \

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$n\pi + \frac{\pi }{4}$

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(A) Given $\sin \left( \frac{\pi }{4} \cot 	heta ight) = \cos \left( \frac{\pi }{4} 	an 	heta ight)$.Using the identity $\cos x = \sin \left( \frac{\pi }{2} - x ight)$,we have $\sin \left( \frac{\pi }{4} \cot 	heta ight) = \sin \left( \frac{\pi }{2} - \frac{\pi }{4} 	an 	heta ight)$.This implies $\frac{\pi }{4} \cot 	heta = \frac{\pi }{2} - \frac{\pi }{4} 	an 	heta$ (considering the principal solution).Dividing by $\frac{\pi }{4}$,we get $\cot 	heta = 2 - 	an 	heta$.Rearranging,$	an 	heta + \cot 	heta = 2$.Since $	an 	heta + \frac{1}{	an 	heta} = 2$,we have $\frac{	an^2 	heta + 1}{	an 	heta} = 2$,which gives $	an^2 	heta - 2 	an 	heta + 1 = 0$.This is $(	an 	heta - 1)^2 = 0$,so $	an 	heta = 1$.Thus,$	heta = n\pi + \frac{\pi }{4}$.

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

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