If sin left(frac{pi}{4} cot thetaright) = cos | 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(\frac{\

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$n \pi + \frac{\pi}{4}, \quad n \in \mathbb{Z}$

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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 get:$\sin \left(\frac{\pi}{4} \cot 	hetaight) = \sin \left(\frac{\pi}{2} - \frac{\pi}{4} 	an 	hetaight)$.This implies $\frac{\pi}{4} \cot 	heta = n \pi + (-1)^n \left(\frac{\pi}{2} - \frac{\pi}{4} 	an 	hetaight)$.For $n=0$,$\frac{\pi}{4} \cot 	heta = \frac{\pi}{2} - \frac{\pi}{4} 	an 	heta \implies \cot 	heta + 	an 	heta = 2 \implies \frac{1}{	an 	heta} + 	an 	heta = 2$.Let $	an 	heta = t$,then $\frac{1}{t} + t = 2 \implies t^2 - 2t + 1 = 0 \implies (t-1)^2 = 0 \implies 	an 	heta = 1$.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 the general solution of $\theta$ is

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