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

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(D) 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}

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

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(D) Given $\sin \left(\frac{\pi}{4} \cot 	hetaight)=\cos \left(\frac{\pi}{4} 	an 	hetaight)$.Using the identity $\cos x = \sin \left(\frac{\pi}{2} - xight)$,$\sin \left(\frac{\pi}{4} \cot 	hetaight) = \sin \left(\frac{\pi}{2} - \frac{\pi}{4} 	an 	hetaight)$.Equating the arguments (general solution $\sin A = \sin B \implies A = n\pi + (-1)^n B$ is complex here,but we check the principal branch),$\frac{\pi}{4} \cot 	heta = \frac{\pi}{2} - \frac{\pi}{4} 	an 	heta$.$\frac{\pi}{4} (\cot 	heta + 	an 	heta) = \frac{\pi}{2}$.$\cot 	heta + 	an 	heta = 2$.$\frac{\cos 	heta}{\sin 	heta} + \frac{\sin 	heta}{\cos 	heta} = 2$.$\frac{\cos^2 	heta + \sin^2 	heta}{\sin 	heta \cos 	heta} = 2$.$\frac{1}{\sin 	heta \cos 	heta} = 2$.$1 = 2 \sin 	heta \cos 	heta$.$1 = \sin 2 	heta$.Since $\sin 2 	heta = 1$,we have $2 	heta = 2n \pi + \frac{\pi}{2}$.Dividing by $2$,we get $	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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