Question

Prove that cotA

/1−cotA

+

tanA

/1−tanA

= − 1 .​

Answer 1

Answer:

Step-by-step explanation:

To Prove :-

$\dfrac{cotA}{1 - cotA} + \dfrac{tanA}{1 - tanA} = - 1$

Solution :-

$\dfrac{cotA}{1 - cotA} + \dfrac{tanA}{1 - tanA} = - 1$

We know that , cotA  = 1/tanA

$\dfrac{\dfrac{1}{tanA}}{1 - \dfrac{1}{tanA}} + \dfrac{tanA}{1 - tanA} = -1$

$\dfrac{\dfrac{1}{tanA}}{\dfrac{tanA - 1}{tanA}} + \dfrac{tanA}{1 - tanA} = - 1$

$\dfrac{1}{tanA - 1}+\dfrac{tanA}{1 - tanA} = -1$

Taking "-1" common :-

$\dfrac{1}{tanA - 1} + \dfrac{tanA}{-1(tanA - 1)} = -1$

$\dfrac{1}{tanA - 1} - \dfrac{tanA}{tanA - 1} = -1$

$\dfrac{1 - tanA}{tanA - 1} = -1$

Taking "-1" common :-

$\dfrac{-1(tanA - 1)}{tanA - 1} = -1$

$-1 = -1$

Hence , L.H.S = R.H.S