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# Prove the following identities where the angles involved are acute angles for which the expressions are defined (cosec theta - cot theta)^2 =1-cos theta/1 + cos theta

5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

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

Views

We need to prove-
$(\csc \theta -\cot \theta )^{2}= \frac{1-\cos \theta }{1+\cos \theta }$

Now, taking LHS,

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

LHS = RHS

Hence proved.

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