Question

$\huge\color{pink}\mathfrak{Answer}$

$\sf\color{red}{prove \: that \: \frac{cotA}{1 - cotA} + \frac{tanA}{1 - tanA} = - 1}$





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Answer 1

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Answer 2

Answer:

Step-by-step explanation:

To Prove :-

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

Solution :-

Taking L.H.S :-

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

We know that :- cotA = 1\tanA

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

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

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

Taking "-" common :-

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

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

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

Taking "-" common :-

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

= -1

Hence Proved