Question

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Answer 1

Step-by-step explanation:

Question:-

$\sf Prove \: that:-$

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

Solution:-

$\sf LHS = \sqrt{ \dfrac{sec \theta - 1}{sec \theta + 1} } + \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} }$

$\sf By \: rationalizing \: the \: denominator$

$\sf = \sqrt{ \dfrac{sec \theta - 1}{sec \theta + 1} \times \dfrac{sec \theta - 1}{sec \theta - 1}} + \sqrt{ \dfrac{sec \theta + 1}{sec \theta - 1} \times \dfrac{sec \theta + 1}{sec \theta + 1} }$

$\sf = \sqrt{ \dfrac{(sec \theta - 1)(sec \theta - 1)}{(sec \theta + 1)(sec \theta - 1)} } + \sqrt{ \dfrac{(sec \theta + 1)(sec \theta + 1)}{(sec \theta - 1)(sec \theta + 1)} }$

$\sf = \sqrt{ \dfrac{(sec \theta - 1) ^{2} }{sec ^{2} \theta - 1} } + \sqrt{ \dfrac{(sec\theta + 1)^{2} }{sec^{2} \theta - 1} }$

$\sf = \dfrac{ \sqrt{(sec \theta - 1) ^{2} }}{ \sqrt{tan ^{2} \theta }} + \dfrac{ \sqrt{( sec\theta + 1)^{2}} }{ \sqrt{tan^{2} \theta} } \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg( \because {sec}^{2} \theta - 1 = {tan}^{2} \theta \bigg)$

$\sf = \dfrac{ sec \theta - 1}{ tan \theta } + \dfrac{ sec\theta + 1 }{ tan\theta}$

$\sf = \dfrac{ sec \theta - 1 + sec\theta + 1 }{ tan \theta }$

$\sf = \dfrac{ 2 sec\theta }{ tan \theta }$

$\sf = \dfrac{ 2 \times \dfrac{1}{cos \theta} }{ \dfrac{sin \theta}{cos \theta} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg(\because sec\theta = \dfrac{1}{cos\theta} \: , \: tan\theta = \dfrac{sin\theta}{cos\theta}\bigg)$

$\sf = \dfrac{ 2 }{ sin \theta} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg(\because \dfrac{1}{sin\theta} = cosec\theta\bigg)$

$\sf =2cosec \theta = RHS$

LHS = RHS

Hence Proved