Question

Help me please with this question.​

Answer 1

$\large\underline{\sf{To\:Prove - }}$

$\rm :\longmapsto\:\dfrac{1}{sec\theta - tan\theta } + \dfrac{1}{sec\theta + tan\theta } = 2sec\theta$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\rm :\longmapsto\:(x + y)(x - y) = {x}^{2} - {y}^{2}$

$\rm :\longmapsto\: {sec}^{2}\theta - {tan}^{2} \theta = 1$

$\large\underline{\sf{Solution-}}$

Consider, LHS

$\rm :\longmapsto\:\dfrac{1}{sec\theta - tan\theta } + \dfrac{1}{sec\theta + tan\theta }$

$\rm \:  =  \:  \: \dfrac{sec\theta + tan\theta + sec\theta - tan\theta }{(sec\theta - tan\theta )(sec\theta + tan\theta )}$

$\rm \:  =  \:  \: \dfrac{2sec\theta }{ {sec}^{2} \theta - {tan}^{2} \theta }$

$\rm \:  =  \:  \: \dfrac{2sec\theta }{1}$

$\rm \:  =  \:  \: 2sec\theta$

Hence,

$\rm :\longmapsto\:\dfrac{1}{sec\theta - tan\theta } + \dfrac{1}{sec\theta + tan\theta } = 2sec\theta$

Additional Information :-

Let's solve one more problem of same type!!

To prove :-

$\rm :\longmapsto\:\dfrac{1}{1 - sin\theta } + \dfrac{1}{1 + sin\theta } = 2 {sec}^{2}\theta$

Identities Used :-

$\rm :\longmapsto\:(x + y)(x - y) = {x}^{2} - {y}^{2}$

$\rm :\longmapsto\: {sin}^{2}\theta + {cos}^{2}\theta = 1$

Solution :-

Consider LHS

$\rm :\longmapsto\:\dfrac{1}{1 - sin\theta } + \dfrac{1}{1 + sin\theta }$

$\rm \:  =  \:  \: \dfrac{1 + sin\theta + 1 - sin\theta }{(1 - sin\theta )(1 + sin\theta )}$

$\rm \:  =  \:  \: \dfrac{2}{1 - {sin}^{2}\theta }$

$\rm \:  =  \:  \: \dfrac{2}{{cos}^{2}\theta }$

$\rm \:  =  \:  \: {2sec}^{2} \theta$

Hence,

$\rm :\longmapsto\:\dfrac{1}{1 - sin\theta } + \dfrac{1}{1 + sin\theta } = 2 {sec}^{2}\theta$

Let's solve one more problem now!!

To prove :-

$\rm :\longmapsto\: \sqrt{\bigg(\dfrac{cosec\theta + 1}{cosec\theta - 1} \bigg) } - \sqrt{\bigg(\dfrac{cosec\theta - 1}{cosec\theta + 1} \bigg) }$

Identities Used :-

$\rm :\longmapsto\: {cosec}^{2}\theta - {cot}^{2}\theta = 1$

$\rm :\longmapsto\:cot\theta = \dfrac{1}{tan\theta }$

Solution :-

$\rm :\longmapsto\: \sqrt{\bigg(\dfrac{cosec\theta + 1}{cosec\theta - 1} \bigg) } - \sqrt{\bigg(\dfrac{cosec\theta - 1}{cosec\theta + 1} \bigg) } = 2tan\theta$

$\rm \:  =  \:  \: \dfrac{cosec\theta + 1 - (cosec\theta - 1)}{ \sqrt{(cosec\theta - 1)(cosec\theta + 1)} }$

$\rm \:  =  \:  \: \dfrac{cosec\theta + 1 - cosec\theta + 1}{ \sqrt{ {cosec}^{2} \theta - {(1)}^{2} } }$

$\rm \:  =  \:  \: \dfrac{2}{ \sqrt{ {cosec}^{2} \theta - 1} }$

$\rm \:  =  \:  \: \dfrac{2}{ \sqrt{ {cot}^{2} \theta} }$

$\rm \:  =  \:  \: \dfrac{2}{cot\theta }$

$\rm \:  =  \:  \: 2tan\theta$

Hence,

$\rm :\longmapsto\: \sqrt{\bigg(\dfrac{cosec\theta + 1}{cosec\theta - 1} \bigg) } - \sqrt{\bigg(\dfrac{cosec\theta - 1}{cosec\theta + 1} \bigg) } = 2tan\theta$