Question

Prove :

$\frac{ \tan(θ) }{1 - \cot(θ) } + \frac{ \cot(θ) }{1 - \tan(θ) } = 1 + \tan(θ) + \cot(θ)$
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Answer 1

Answer:

$\huge \mathfrak {Question:-}$

Prove:-

$\frac{ \tan(θ) }{1 - \cot(θ) } + \frac{ \cot(θ) }{1 - \tan(θ) } + 1 + \tan(θ) + \cot(θ)$

$\huge \mathfrak {Solution : - }$

$\bf L.H.S - \frac{ \tanθ }{1 - \cotθ } + \frac{ \cotθ }{1 - \tanθ } = \frac{ { - \tan}^{2} θ }{1 - \tanθ } + \frac{ \cotθ}{1 - \tanθ } = \frac{ - { \tan }^{2} θ + \cotθ }{1 - \tan θ} \\ \bf multiply \: \tanθ \: by \tanθ \: we \: get : \\ \bf = \frac{1 - { \tan }^{3} θ}{ \tanθ -( 1 - \tanθ)} \\ \bf = \frac{(1 - \tanθ)(1 + \tanθ + { \tan }^{2} θ)}{ \tanθ(1 - \tanθ)} \\ \bf = \frac{(1 + \tanθ + { \tan }^{2}θ) }{ \tanθ} \\ \bf = \cot θ + 1 + \tanθ = R.H.S$

Hence Proved !!

Answer 2

Answer:

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