Question

Prove that:

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Answer 1

Step-by-step explanation:

Question:-

$\rm Prove \: that:-$

$\rm \dfrac{sin \theta}{cot \theta + cosec \theta} = 2 + \dfrac{sin \theta}{cot \theta - cosec \theta}$

Proof:-

Let us Prove by simplifying LHS and RHS seperately.

Taking LHS:-

$\rm LHS = \dfrac{sin \theta}{cot \theta + cosec \theta}$

$\rm = \dfrac{sin \theta}{ \dfrac{cos \theta}{sin \theta} + \dfrac{1}{sin \theta} }$

$\rm = \dfrac{sin \theta}{ \dfrac{cos \theta + 1}{sin \theta} }$

$\rm = \dfrac{sin \theta \times sin \theta}{ cos \theta + 1 } = \dfrac{sin ^{2} \theta}{ cos \theta + 1}$

$\rm = \dfrac{1 - {cos}^{2} \theta }{ cos \theta + 1 } \: \: \: \: \: \: \: \: \: \: \: \: \: \bigg ( \because \: sin ^{2} = 1 - {cos}^{2} \theta \bigg)$

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

$\boxed{ \rm \leadsto 1 - {cos} \theta = LHS}$

Now, Taking RHS

$\rm RHS = 2 + \dfrac{sin \theta}{cot \theta - cosec \theta}$

$\rm = 2 + \dfrac{sin \theta}{ \dfrac{cos \theta}{sin \theta} - \dfrac{1}{sin \theta} }$

$\rm = 2+ \dfrac{sin \theta}{ \dfrac{cos \theta - 1}{sin \theta} }$

$\rm =2 + \dfrac{sin \theta \times sin \theta}{ cos \theta - 1 } = 2 + \dfrac{sin ^{2} \theta}{ cos \theta - 1}$

$\rm =2+ \dfrac{1 - {cos}^{2} \theta }{ -(cos \theta - 1) }$

$\rm = 2- \dfrac{(1 - {cos} \theta )(1 + cos \theta)}{ 1 - cos\theta }$

$\rm = 2-(1 + cos \theta)$

$\rm = 2 - 1 - cos\theta$

$\boxed{ \rm \leadsto 1 - {cos} \theta = RHS}$

$\boxed{ \rm \dashrightarrow LHS = RHS = 1 - {cos} \theta }$

LHS = RHS

Hence Proved