If 2 sec 2alpha = tan beta + cot beta,then on | vedclass

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(A) The given equation is $2 \sec 2\alpha = 	an \beta + \cot \beta$.We can rewrite the right side as:$	an \beta + \cot \beta = \frac{\sin \beta}{\co

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$\frac{\pi}{4}$

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(A) The given equation is $2 \sec 2\alpha = 	an \beta + \cot \beta$.We can rewrite the right side as:$	an \beta + \cot \beta = \frac{\sin \beta}{\cos \beta} + \frac{\cos \beta}{\sin \beta} = \frac{\sin^2 \beta + \cos^2 \beta}{\sin \beta \cos \beta} = \frac{1}{\sin \beta \cos \beta}$.Multiplying numerator and denominator by $2$,we get $\frac{2}{2 \sin \beta \cos \beta} = \frac{2}{\sin 2\beta}$.So,the equation becomes $2 \sec 2\alpha = \frac{2}{\sin 2\beta}$,which simplifies to $\frac{2}{\cos 2\alpha} = \frac{2}{\sin 2\beta}$.This implies $\cos 2\alpha = \sin 2\beta$.Using the identity $\sin 	heta = \cos(\frac{\pi}{2} - 	heta)$,we have $\cos 2\alpha = \cos(\frac{\pi}{2} - 2\beta)$.Therefore,$2\alpha = \frac{\pi}{2} - 2\beta$,which gives $2\alpha + 2\beta = \frac{\pi}{2}$.Dividing by $2$,we get $\alpha + \beta = \frac{\pi}{4}$.

If $2 \sec 2\alpha = \tan \beta + \cot \beta$,then one of the values of $\alpha + \beta$ is

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