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

Correct Answer is : cotα(log|sinx|−log|sin(x+α)|)+C

\tan \alpha(\log (\sec (x+a))+\log \sec x+C)

\cot \alpha(\log |\sin x|-\log |\sin (x+\alpha)|)+C

\tan \alpha(\log (\frac{\cos x}{\sin (x+\alpha)}))+C

\cot \alpha(\log \frac{\sin (x+a)}{\cos (x+\alpha)})+C

AP EAPCET 04-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

∫‌

1+tan‌x‌tan(x+a)

tan‌x‌tan(x+a)

dx=

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