Question

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Answer 1

Answer is given below

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Answer 2

Question :

Prove that :

$\sqrt{ \frac{sec \theta - 1}{sec \theta + 1} } + \sqrt{ \frac{sec \theta + 1}{sec \theta - 1} } = 2cosec \theta$

Proof :

LHS=

$\implies {\small{\sqrt { \frac {(sec \theta - 1 )( sec \ theta - 1)}{(sec \theta + 1)(sec \theta - 1)} }} + \sqrt{ \frac{(sec \theta + 1)(sec \theta + 1)}{(sec \theta - 1)(sec \theta + 1)} } }$

$\implies \sqrt{ \frac{ {(sec \theta - 1)}^{2} }{ {sec \theta }^{2} - 1 } } + \sqrt{ \frac{ {(sec \theta + 1)}^{2} }{{sec \theta }^{2} - 1 } } \\ \implies \sqrt{ \frac{{(sec \theta - 1)}^{2}}{ {tan}^{2} \theta} } + \sqrt{ \frac{{(sec \theta + 1)}^{2}}{{tan}^{2} \theta} } \\ \implies \frac{sec \theta - 1}{tan \theta} + \frac{sec \theta + 1}{tan \theta} \\ \implies \frac{ sec \theta \cancel { - 1 } + sec \theta \cancel{ + 1}}{tan \theta} \\ \implies \frac{2sec \theta}{tan \theta} \\ \implies \frac{2}{ \cancel{cos \theta}} \times \frac{ \cancel{cos \theta}}{sin \theta} \\ \implies 2cosec \theta$=RHS

Identities Used :

$1. {sec}^{2} \theta - 1 = {tan}^{2} \theta \\ 2.(a + b)(a - b) = {a}^{2} - {b}^{2} \\ 3. sec \theta = \frac{1}{cos \theta} \\ 4.tan \theta = \frac{sin \theta}{cos \theta}\\4.\frac{1}{sin \theta}=cosec \theta$