Question

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Answer 1

To Prove :

$\tt\frac{1 + \cos \: A}{1 - \cos \: A} = \frac{ { \tan }^{2}A}{ { (\sec A- 1) }^{2} }$

Solution :

Let's try to Prove RHS = LHS.

Taking RHS

$\tt \dashrightarrow \frac{ {\tan }^{2} A }{ { (\sec A - 1)}^{2} }$

$\tt \dashrightarrow \frac{ ( { \sec }^{2} A - 1) }{{( \sec A- 1 ) }^{2} }$

$\tt \dashrightarrow \frac{( \sec A + 1)( \sec A - 1)}{( \sec A - 1)( \sec a - 1)}$

$\tt \dashrightarrow \frac{( \sec A + 1)}{( \sec A - 1)}$

$\tt \dashrightarrow \frac{ \frac{1 + \cos A}{ \cos A} }{ \frac{1 - \cos A}{ \cos A}}$

$\tt \dashrightarrow \frac{1 + \cos \: A}{1 - \cos \: A}$

Hence Proved.

Other method, Provide in attachment.

$\rule{200}3$

$\boxed{\begin{minipage}{7 cm}Fundamental Trigonometric Identities \\ \\ \sin^2\theta + \cos^2\theta=1 \\ \\ 1+\tan^2\theta = \sec^2\theta \\ \\ 1 + \cot^2\theta = \text{cosec}^2 \, \theta </p><p>\end{minipage}}$

Answer 2

SoluTion:

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Given Expression :

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$\sf{\dfrac{1\:+\:CosA}{1\:-\:CosA}\:=\:\dfrac{Tan^{2} A}{(SecA\:-\:1)^2}}$

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Taking RHS,

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$\sf{\dfrac{Tan^{2} A}{(SecA\:-\:1)^2}}$

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We know that,

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$\large{\boxed{\red{\sf{Tan^{2} A\:=\:Sec^{2} A\:-\:1}}}}$

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$\longrightarrow$ $\sf{\dfrac{(Sec^{2} A\:-\:1)}{(SecA\:-\:1)(SecA\:-\:1)}}$

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We also know that,

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$\boxed{\green{\sf{(a^2\:-\:b^2\:)\:=\:(a\:+\:b)(a\:-\:b)}}}$

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$\longrightarrow$ $\sf{\dfrac{(SecA\:-\:1)(SecA\:+\:1)}{(SecA\:-\:1)(SecA\:-\:1)}}$

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Cancelling SecA - 1 from both numerator and denominator,

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$\longrightarrow$ $\sf{\dfrac{SecA\:+\:1}{SecA\:-\:1}}$

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Converting sec into cos,

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$\longrightarrow$ $\sf{\dfrac{ \dfrac{1}{cosA} \: + \: \dfrac{cosA}{cosA} }{\dfrac{1}{cosA} \: - \: \dfrac{cosA}{cosA} }}$

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Taking LCM, we get,

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$\longrightarrow$ $\sf{\blue{\dfrac{1\:+\:CosA}{1\:-\:CosA}}}$

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$\longrightarrow$ LHS

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Hence Proved!