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Answer»

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 \\ \\ $\sin^2\theta + \cos^2\theta=1 \\ \\ 1+\tan^2\theta = \sec^2\theta \\ \\ 1 + \cot^2\theta = \text{cosec}^2 \, \theta$\end{minipage}}


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