Question

√secA-1/secA+1+√secA-1/secA+1=2cosecA

Answer 1

Answer:

your answer attached in the photo

Answer 2

$: \longrightarrow \underline{ \underline{\mathfrak{ \: Required \: \: \: Answer :-}}}$

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$\tt {\red{ : \implies \sqrt{ \frac{sec \: A - 1}{sec \:A + 1 } } + \sqrt{ \frac{sec \: A + 1}{sec \: A - 1} } = 2 \: cosec \: A}}$

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$\tt{ \:L.H.S = } \tt{ \green{ \sqrt{ \frac{sec \:A - 1 }{sec \: A + 1} } + \sqrt{ \frac{sec \: A + 1}{sec \: A - 1} } }} \\ \\ \sf \: On \: squaring \: to \: remove \: the \: roots, \\ \\ : \longrightarrow \tt{ \green{ \: \sqrt{ \frac{(sec \:A - 1) {}^{2} }{(sec \: A + 1)(sec \:A - 1) } } \: \: + \: \: \sqrt{ \frac{ {(sec \:A + 1) }^{2} }{(sec \: A - 1)(sec \: A + 1)} } }} \\ \\ : \longrightarrow \tt{ \green{ \: \sqrt{ \frac{( {sec \: A - 1)}^{ \cancel2} }{ {sec}^{ \cancel2} A - 1} } + \sqrt{ \frac{( {sec \: A + 1)}^{ \cancel2} }{ {sec}^{ \cancel2} A - 1} } }} \\ \\ : \longrightarrow \tt{ \green{ \: \frac{sec \: A - 1}{sec \: A - 1} + \frac{sec \:A + 1 }{sec \:A - 1 } }} \\ \\ : \longrightarrow \tt{ \green{ \: \frac{(sec \: A - 1) + (sec \: A + 1)}{ sec \: A - 1}}} \\ \\ : \longrightarrow \tt{ \green{ \frac{sec \:A - \cancel1 + sec \: A + \cancel1}{sec \: A - 1} }} \\ \\ : \longrightarrow \tt{ \green{ \: \frac{2 \: sec \: A}{tan \: A} }} \\ \\ : \longrightarrow \tt{ \green{ \: \frac{ \frac{2}{ \cancel{cos \: A}} }{ \frac{sin \: A}{ \cancel{cos \:A} } } }} \\ \\ : \longrightarrow \tt{ \green{ \: \frac{2}{sin \: A} }} = 2 \: cosec \: A \\ \\ \tt{ \therefore \: L.H.S = R.H.S}$

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@MissSolitary

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