Question

Prove , ....kindly send answer soon crct answer ll be marked as BRAINILIST ​

Answer 1

Answer:

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jai hind

Answer 2

Prove the following :-

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

Solution

$\sf L.H.S. = \dfrac{1 - \cos \theta}{1 + \cos \theta} \\\\ \qquad\sf [Multiplying \: (1 - \cos \theta) \: in\: both \: numerator \: \& \: denominator] \\\\ \implies\sf \dfrac{(1 - \cos \theta)^2}{(1 + \cos\theta)(1 - \cos \theta)} \\\\ \implies\sf \dfrac{(1 - \cos\theta)^2}{1 - \cos^2 \theta} \\\\ \implies\sf \dfrac{(1 - \cos \theta)^2}{\sin^2 \theta} \qquad [\because 1 - \cos^2 \theta = \sin^2 \theta] \\\\ \implies\sf \bigg(\dfrac{1 - \cos \theta}{\sin\theta} \bigg)^2 \\\\ \implies\sf \bigg(\dfrac{1}{\sin \theta} - \dfrac{\cos \theta}{\sin \theta} \bigg)^2 \\\\ \qquad\sf [Identity : \dfrac{1}{\sin \theta} = cosec \theta \: ; \: \dfrac{\cos\theta}{\sin \theta} = \cot \theta] \\\\ \implies\sf\boxed{ (cosec \theta - \cot \theta)^2 } \\\\ \implies\sf R.H.S.$

Therefore,

$\bold{ \dfrac{1 - cos \theta}{1 + cos \theta} \: =\: (cosec \theta - cot \theta)^2 \qquad [Hence\: proved] }$