int frac{1 + tan x tan(x + a)}{tan x tan(x + | vedclass

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(B) Let $I = \int \frac{1 + 	an x 	an(x + a)}{	an x 	an(x + a)} dx$. We know that $	an a = 	an((x + a) - x) = \frac{	an(x + a) - 	an x}{1 +

Vedclass · Accepted Answer

$\cot a (\log |\sin x| - \log |\sin(x + a)|) + C$

Vedclass · Answer

(B) Let $I = \int \frac{1 + 	an x 	an(x + a)}{	an x 	an(x + a)} dx$. We know that $	an a = 	an((x + a) - x) = \frac{	an(x + a) - 	an x}{1 + 	an x 	an(x + a)}$. Therefore,$1 + 	an x 	an(x + a) = \frac{	an(x + a) - 	an x}{	an a}$. Substituting this into the integral,we get: $I = \int \frac{	an(x + a) - 	an x}{	an a 	an x 	an(x + a)} dx$. $I = \frac{1}{	an a} \int \left( \frac{	an(x + a)}{	an x 	an(x + a)} - \frac{	an x}{	an x 	an(x + a)} ight) dx$. $I = \cot a \int \left( \frac{1}{	an x} - \frac{1}{	an(x + a)} ight) dx$. $I = \cot a \int (\cot x - \cot(x + a)) dx$. Integrating,we get: $I = \cot a (\log |\sin x| - \log |\sin(x + a)|) + C$. Thus,the correct option is $B$.

$\int \frac{1 + \tan x \tan(x + a)}{\tan x \tan(x + a)} dx =$

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