Question

Prove the given identity :
(The angles involved are acute angles )
$( \cosec θ- cotθ) ^{2} = \frac{1 - \cosθ}{1 + \cosθ}$
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Answer 1

Question:

Prove the given identity:

$\sf{(cosec \theta - cot \theta)^{2}=\dfrac{1-cos \theta}{1+cos \theta}}$

Procedure:

• First, we will convert the cosec θ and cot θ in the RHS to sin θ and cos θ.
• This step is done because there is only cos θ on RHS and not cosec θ or cot θ.

Solution:

Taking LHS,

$\sf{(cosec \theta - cot \theta)^{2}}$

$:\implies \: \sf{\bigg(\dfrac{1}{sin \theta}-\dfrac{cos \theta}{sin \theta}\bigg)^{2}}$

$\longrightarrow \: \: \sf{\bigg(\dfrac{1-cos \theta}{sin \theta}\bigg)^{2}}$

$\longrightarrow \: \: \sf{\dfrac{(1-cos \theta)^{2}}{(sin \theta)^{2}}}$

$\hookrightarrow \: \: \sf{\dfrac{(1-cos \theta)^{2}}{sin^{2}\theta}}$

sin²θ can also be written as (1 - cos²θ)

$\implies \: \sf{\dfrac{(1-cos \theta)(1-cos \theta)}{(1-cos^{2}\theta)}}$

Using the identity (a + b) (a - b) = a² - b²,

$\implies \: \sf{\dfrac{(1-cos \theta)(1-cos \theta)}{(1-cos\theta)(1+cos \theta)}}$

$\implies \: \sf{\dfrac{\cancel{(1-cos \theta)}(1-cos \theta)}{\cancel{(1-cos\theta)}(1+cos \theta)}}$

$\leadsto \: \: \sf{\dfrac{1-cos \theta}{1+cos \theta}}$

$\therefore \: \boxed{\bf{(cosec \theta - cot \theta)^{2}=\dfrac{1-cos \theta}{1+cos \theta}}}$

Answer 2

Answer:-

LHS = (cosecθ - cotθ)²

= ${( \frac{1}{sinθ} - \frac{ \cosθ }{ \sinθ} ) }^{2}$

$= { \frac{(1 - \cosθ)}{1 - \cosθ} }^{2} = \frac{1 - \cos θ}{1 + \cosθ }$

= RHS