Question

$\int 1+\tan x.\tan (x-a)dx=\cot a\log \frac{\cos x}{\cos (x-a)}$. If this is true enter 1, else enter 0.

Answer 1

$I=\int 1+\tan x.\tan (x-a)dx=\int \frac{\cos x\cos (x-a)+\sin x\sin (x-a)}{\cos x\cos (x-a)}dx$
$=\int \frac{\cos a}{\cos x\cos (x-a)}dx=\int (\frac{\cos a}{\sin a}.\frac{\sin a}{\cos x\cos (x-a)})dx$
$=\int (\frac{\cos a}{\sin a}.\frac{\sin [x-(x-a)]}{\cos x\cos (x-a)})dx=\int \cot a[\tan x-\tan (x-a)]dx$
$=\cot a\log \left|\frac{\sec x}{\sec (x-a)}\right|$