Question

Prove that :
${( \csc\theta - \cot\theta)}^{2} = \frac{1 - \cos\theta}{1 + \cos\theta}$


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Answer 1

Answer:

hello

Step-by-step explanation:

$( \csc(a) - \cot(a) ) {}^{2} = \frac{1 - \cos( \alpha ) }{1 + \cos( \alpha ) } \\ = ( \frac{1}{ \sin( \alpha ) } - \frac{ \cos( \alpha ) }{ \sin( \alpha ) } ) {}^{2} \\ = ( \frac{1 - \cos( \alpha ) }{ \sin( \alpha ) } ) {}^{2} \\ = \frac{(1 - \cos( \alpha )) }{ \sin {}^{2} ( \alpha ) } \\ = \frac{(1 - \cos( \alpha) ) (1 - \cos( \alpha ) )}{(1 + \cos( \alpha ) )(1 - \cos( \alpha )) } \\ = \frac{1 - \cos( \alpha ) }{1 + \cos( \alpha ) }$

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Answer 2

ANSWER:

To Prove:

$\:\:\:\bullet\:\:\:(\csc\theta-\cot\theta)^2=\dfrac{1-\cos\theta}{1+\cos\theta}$

Proof:

We need to that,

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

Solving LHS,

$\implies(\csc\theta-\cot\theta)^2$

We know that,

$\hookrightarrow\csc\theta=\dfrac{1}{\sin\theta}$

And,

$\hookrightarrow\cot\theta=\dfrac{\cos\theta}{\sin\theta}$

So,

$\implies(\csc\theta-\cot\theta)^2$

Hence,

$\implies\left(\dfrac{1}{\sin\theta}-\dfrac{\cos\theta}{\sin\theta}\right)^2$

On simplifying,

$\implies\left(\dfrac{1-\cos\theta}{\sin\theta}\right)^2$

We can write the above mentioned expression as,

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

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

We know that,

$\hookrightarrow\sin^2\theta=1-\cos^2\theta$

So,

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

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

We know that,

$\hookrightarrow a^2-b^2=(a+b)(a-b)$

So,

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

On simplifying,

$\implies\bf\dfrac{1-cos\theta}{1+cos\theta}=RHS$

HENCE PROVED!!!