Question

prove that tanA/1-cotA+cotA/1-tanA=1+secA cosecA​

Answer 1

Step-by-step explanation:

very important question

Answer 2

ANSWER:-

Given:

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

To prove:

1 + secA cosecA

Proof:

Take L.H.S

$\frac{tan \: A}{1 - cot \: A} + \frac{cot \: A}{1 - tan \: A} \\ \\ = > \frac{tan \: A}{1 - \frac{1}{tan \: A} } + \frac{ \frac{1}{tan \: A} }{1 - tan \: A} \\ \\ = > \frac{ \frac{tan \: A}{tan \: A - 1} }{tan \: A} + \frac{1}{tan \: A(1 - tan \: A)} \\ \\ = > \frac{ {tan}^{2} A}{tan \: A - 1} + \frac{1}{tan \: A(1 - tan \: A)} \\ \\ = > \frac{ - {tan}^{2} A}{1 - tan \: A} + \frac{1}{tan \: A(1 - tan \: A)} \\ \\ = > \frac{ - {tan}^{3} A + 1}{tan \: A(1 - tan \: A)} \\ \\ = > \frac{ {1}^{3} - {tan}^{3} A }{tan \: A(1 - tan \: A)} \\ \\ = > \frac{(1 - tan \: A)(1 + {tan}^{2} A + tan \: A) }{tan \: A(1 - tan \: A)} \\ \\ = > \frac{(1 + { tan }^{2} A + tan \: A)}{tan \: A} \\ \\ = > \frac{( {sec}^{2} A + tan \: A) }{tan \: A} \\ \\ = > \frac{ {sec}^{2} A}{tan \: A} + \frac{tan \: A}{tan \: A} \\ \\ = > {sec}^{2} A \: cot \: A + 1 \\ \\ = > \frac{1}{ {cos}^{2} A} \times \frac{cos \: A}{sin \: A} + 1 \\ \\ = > \frac{1}{cos \: A sin \: A} + 1 \\ \\ = > 1 + sec \: A \: cosec \: A \: \: \: \: \: \: [R.H.S.]$

Hence,

Proved.

Hope it helps ☺️