Question

plz someone answer this quickly ​

Answer 1

Question :–

$\\ \: \large \to \: { \bold{show \: \: that : \frac{ \sin( \theta) }{1 + \cos( \theta) } + \frac{1 + \cos( \theta) }{ \sin( \theta) } = 2cosec( \theta)}}$

ANSWER :–

TO PROVE :–

$\\ \: { \bold{ \dfrac{ \sin( \theta) }{1 + \cos( \theta) } + \dfrac{1 + \cos( \theta) }{ \sin( \theta) } = 2cosec( \theta)}}$

SOLUTION :–

• Let's take L.H.S. –

$\\ \: { \bold{ = \dfrac{ \sin( \theta) }{1 + \cos( \theta) } + \dfrac{1 + \cos( \theta) }{ \sin( \theta) } }}$

• We should write this as –

$\\ \: { \bold{ = \dfrac{[ 1 + \cos( \theta)] ^{2} + { \sin }^{2} ( \theta)}{ \sin( \theta)[1 + \cos( \theta) ]} }}$

$\\ \: { \bold{ = \dfrac{1 + \cos ^{2} ( \theta) + 2 \cos( \theta) + { \sin }^{2} ( \theta)}{ \sin( \theta)[1 + \cos( \theta) ]} \: \: \large [ \: \because \: \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \: ] \: \: }}$

$\\ \: { \bold{ = \dfrac{1 + 1 + 2 \cos( \theta) }{ \sin( \theta)[1 + \cos( \theta) ]} \: \large [ \: \because \: \: {sin}^{2} \theta + {cos}^{2} \: \theta = 1\: ] \: \:}}$

$\\ \: { \bold{ = \dfrac{2+ 2 \cos( \theta) }{ \sin( \theta)[1 + \cos( \theta) ]} }}$

$\\ \: { \bold{ = \dfrac{2[1 + \cos( \theta) ] }{ \sin( \theta)[1 + \cos( \theta) ]} }}$

$\\ \: { \bold{ = \dfrac{2 }{ \sin( \theta)} \: \: \large [ \: \because \: \: \dfrac{1}{ \sin( \theta) } \: \: = cosec (\theta) ] \: \: }}$

$\\ \: { \bold{ = 2 \: cosec( \theta) }}$

$\\ \: { \bold{ = R.H.S. \: \: \: \: \: \: [Hence \: \: proved]}}$

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