Discussion Forum

The expression   $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$   can be written as

Answer:

Explanation:

Given expression is 

$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$

 = $\frac{\sin A}{\cos A}\times\frac{\sin A}{\sin A-\cos A}+\frac{\cos A}{\sin A}\times\frac{\cos A}{\cos A-\sin A}$

     = $\frac{1}{\sin A-\cos A}\left\{\frac{\sin^{3}A-\cos^{3} A}{\cos A\sin A}\right\}$

   = $\frac{\sin^{2}A+\sin A \cos A+\cos^{2} A}{\cos A\sin A}$

     =   $\frac{1-\sin A \cos A}{\cos A\sin A}$

       =1+ sec A cosec A