Question

$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\sec AcosecA+1$.

Answer 1

$\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}=\sec A\text{cosec}A+1$

$\underset{–}{L.H.S}$

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

$=\frac{{\tan }^{2}A}{\tan A-1}+\frac{1}{\tan A(1-\tan A)}=\frac{1}{\tan A-1}({\tan }^{2}A-\frac{1}{\tan A})=\frac{1}{\tan A-1}\left(\frac{{\tan }^{3}A-1}{\tan A}\right)\therefore \frac{1}{\tan A-1}(\tan A-1)\left(\frac{{\tan }^{2}A+1+\tan A}{\tan A}\right)$

$=\frac{{\sec }^{2}A+\tan A}{\tan A}(\because 1+{\tan }^{2}A={\sec }^{2}A)=\frac{{\sec }^{2}A}{\tan A}+1=\frac{\frac{1}{{\cos }^{2}A}}{\frac{\sin A}{\cos A}}+1=\frac{1}{\sin A\cos A}+1=\sec AcosecA+1(R.H.S)$