If int tan (x - alpha) cdot tan (x + alpha) c | vedclass

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(A) We have,$	an 2 x = 	an ((x - \alpha) + (x + \alpha))$.Using the identity $	an (A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get:$	an

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$\frac{-3}{2}$

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(A) We have,$	an 2 x = 	an ((x - \alpha) + (x + \alpha))$.Using the identity $	an (A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we get:$	an 2 x = \frac{	an (x - \alpha) + 	an (x + \alpha)}{1 - 	an (x - \alpha) 	an (x + \alpha)}$.Rearranging the terms:$	an 2 x (1 - 	an (x - \alpha) 	an (x + \alpha)) = 	an (x - \alpha) + 	an (x + \alpha)$.$	an 2 x - 	an (x - \alpha) 	an (x + \alpha) 	an 2 x = 	an (x - \alpha) + 	an (x + \alpha)$.Thus,$	an (x - \alpha) 	an (x + \alpha) 	an 2 x = 	an 2 x - 	an (x - \alpha) - 	an (x + \alpha)$.Now,integrating both sides:$\int 	an (x - \alpha) 	an (x + \alpha) 	an 2 x \ d x = \int (	an 2 x - 	an (x - \alpha) - 	an (x + \alpha)) \ d x$.$= \frac{1}{2} \log |\sec 2 x| - \log |\sec (x - \alpha)| - \log |\sec (x + \alpha)| + C$.Comparing this with $p \log |\sec 2 x| + q \log |\sec (x + \alpha)| + r \log |\sec (x - \alpha)| + c$,we get:$p = \frac{1}{2}$,$q = -1$,and $r = -1$.Therefore,$p + q + r = \frac{1}{2} - 1 - 1 = \frac{1}{2} - 2 = \frac{-3}{2}$.

If $\int \tan (x - \alpha) \cdot \tan (x + \alpha) \cdot \tan 2 x \ d x = p \log |\sec 2 x| + q \log |\sec (x + \alpha)| + r \log |\sec (x - \alpha)| + c$,then $p + q + r = . . . . . .$

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