Question

$\frac{tanA}{1-cotA}+\frac{cotA}{1-tanA}=1+secAcosecA$

Answer 1

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

Taking L.H.S.-

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

$=\frac{\tan A}{1-\left(\frac{1}{\tan A}\right)}+\frac{\left(\frac{1}{\tan A}\right)}{1-\tan A}$

$=\frac{{\tan }^{2}A}{\tan A-1}+\frac{1}{\tan A(1-\tan A)}$

$=\frac{1-{\tan }^{3}A}{\tan A(1-\tan A)}$

$=\frac{(1-\tan A)(1+\tan A+{\tan }^{2}A)}{\tan A(1-\tan A)}(\because a^3-b^3=(a-b)(a^2+ab+b^2))$

$=\frac{{\sec }^{2}A+\tan A}{\tan A}(\because 1+{\tan }^{2}A={\sec }^{2}A)$

$=1+\frac{{\sec }^{2}A}{\tan A}$

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

$=1+\sec A\csc A$

$=$ R.H.S.

Hence proved.