Question

Prove that $\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}$= $\mathrm{secA}$$\mathrm{cosecA}+1.$

Answer 1

We have,

$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\sec A\cos ecA+1$

$\begin{array}{c}L.H.S=\frac{\tan A}{1-\frac{1}{\tan A}}+\frac{\frac{1}{\tan A}}{1-\tan A}=\frac{{\tan }^{2}A}{\tan A-1}+\frac{1}{\tan A\left(1-\tan A\right)}\\ =\frac{1}{\tan A-1}\left({\tan }^{2}A-\frac{1}{\tan A}\right)\\ =\frac{1}{\tan A-1}\left(\frac{{\tan }^{3}A-1}{\tan A}\right)\left(\left(a^3+b^3\right)-\left(a-b\right)\left(a^2+b^2+2ab\right)\right)\\ =\frac{1}{\tan A-1}\tan A-1\left(\frac{{\tan }^{2}A+1+\tan A}{\tan A}\right)\\ =\frac{{\sec }^{2}A+\tan A}{\tan A}\left(1+{\tan }^{2}A={\sec }^{2}A\right)\\ =\frac{{\sec }^{2}A}{\tan A}+1=\frac{\frac{1}{{\cos }^{2}A}}{\frac{\sin A}{\cos A}}+1\\ =\frac{1}{\sin A\cos A}+1\\ =\sec A\cos ecA+1\\ L.H.S=R.H.S\end{array}$