If cot theta + cot left( frac{pi }{4} + theta | vedclass

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(D) Given equation: $\cot 	heta + \cot \left( \frac{\pi }{4} + 	heta ight) = 2$ Using $\cot A + \cot B = \frac{\sin(A+B)}{\sin A \sin B}$,we get:

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$n\pi \pm \frac{\pi }{6}$

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(D) Given equation: $\cot 	heta + \cot \left( \frac{\pi }{4} + 	heta ight) = 2$ Using $\cot A + \cot B = \frac{\sin(A+B)}{\sin A \sin B}$,we get: $\frac{\sin(	heta + \frac{\pi}{4} + 	heta)}{\sin 	heta \sin(\frac{\pi}{4} + 	heta)} = 2$ $\frac{\sin(\frac{\pi}{4} + 2	heta)}{\sin 	heta \sin(\frac{\pi}{4} + 	heta)} = 2$ Using $2 \sin A \sin B = \cos(A-B) - \cos(A+B)$: $\sin(\frac{\pi}{4} + 2	heta) = \cos(\frac{\pi}{4} + 	heta - 	heta) - \cos(\frac{\pi}{4} + 	heta + 	heta)$ $\sin(\frac{\pi}{4} + 2	heta) = \cos(\frac{\pi}{4}) - \cos(\frac{\pi}{4} + 2	heta)$ $\sin(\frac{\pi}{4} + 2	heta) + \cos(\frac{\pi}{4} + 2	heta) = \frac{1}{\sqrt{2}}$ Multiply by $\frac{1}{\sqrt{2}}$: $\frac{1}{\sqrt{2}} \sin(\frac{\pi}{4} + 2	heta) + \frac{1}{\sqrt{2}} \cos(\frac{\pi}{4} + 2	heta) = \frac{1}{2}$ $\sin(\frac{\pi}{4} + 2	heta + \frac{\pi}{4}) = \frac{1}{2}$ $\sin(2	heta + \frac{\pi}{2}) = \frac{1}{2}$ $\cos(2	heta) = \frac{1}{2}$ $2	heta = 2n\pi \pm \frac{\pi}{3}$ $	heta = n\pi \pm \frac{\pi}{6}$

If $\cot \theta + \cot \left( \frac{\pi }{4} + \theta \right) = 2$,then the general value of $\theta$ is

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