Question

HELLO MATE!!!!

plz help me in trigonometry :

tanA/1-CotA+CotA/1-tanA=(1+tanA+cotA)​

Answer 1

Step-by-step explanation:

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Answer 2

To prove ---->

$\dfrac{tan \alpha }{1 - cot \alpha } \: + \dfrac{cot \alpha }{1 - tan \alpha } = 1 + tan \alpha + cot \alpha$

Concept used---->

1)

$cot \beta = \dfrac{1}{tan \beta }$

2)

${a}^{3} \: - {b}^{3} \: = (a - b) \: ( {a}^{2} \: + {b}^{2} \: + ab )$

Proof----> LHS

$= \dfrac{tan \alpha }{1 - cot \alpha } + \dfrac{cot \alpha }{1 - tan \alpha } \\ = \dfrac{tan \alpha }{1 - \dfrac{1}{tan \alpha } } + \dfrac{ \dfrac{1}{tan \alpha } }{1 - tan \alpha } \\ = \dfrac{tan \alpha }{ \dfrac{tan \alpha - 1}{tan \alpha } } + \dfrac{1}{ - tan \alpha (tan \alpha - 1)} \\ = \frac{tan ^{2} \alpha }{tan \alpha - 1} - \dfrac{1}{tan \alpha (tan \alpha - 1)}$

$= \dfrac{ {tan}^{3} \alpha - 1 }{tan \alpha \: (tan \alpha \: - \: 1)}$

$= \dfrac{ {(tan \alpha) }^{3} - (1) ^{3} }{tan \alpha \: (tan \alpha - 1)}$

$= \dfrac{(tan \alpha - 1) \: ( {tan}^{2} \alpha + 1 + tan \alpha ) }{tan \alpha \: (tan \alpha - 1)}$

$(tan \alpha - 1) \: is \: cancel \: out \: from \: numerator \: and \: denominator \:$

$\dfrac{ {tan}^{2} \alpha }{tan \alpha } + \dfrac{1}{tan \alpha } + \dfrac{tan \alpha }{tan \alpha }$

$= tan \alpha + cot \alpha + 1$

= RHS