Question

The expression $\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}$ can be written as:

Answer 1

$\begin{array}{c}\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}\\ \Rightarrow \frac{\frac{\sin A}{\cos A}}{1-\frac{\cos A}{\sin A}}+\frac{\frac{\cos A}{\sin A}}{1-\frac{\sin A}{\cos A}}\\ \Rightarrow \frac{{\sin }^{2}A}{\cos A\left(\sin A-\cos A\right)}+\frac{{\cos }^{2}A}{\sin A\left(\sin A-\cos A\right)}\\ \Rightarrow \frac{{\sin }^{3}A-{\cos }^{3}A}{\sin A\cos A\left(\sin A-\cos A\right)}\\ \Rightarrow \frac{\left(1+\sin A\cos A\right)}{\sin A\cos A}\\ \Rightarrow \frac{1}{\sin A.\cos A}+\frac{\sin A\cos A}{\sin A\cos A}\\ \Rightarrow \sec A\cos ecA+1\end{array}$