Question

2+cot^2theta- cosec^2theta /cosec^2 theta -1 =tan^2 theta​

Answer 1

QUESTION:-

Prove that :-

$\sf \dfrac{2 + {cot}^{2} \theta- {cosec}^{2} \theta }{{cosec}^{2} \theta - 1} ={tan}^{2} \theta$

SOLUTION:-

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We have LHS:

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$\implies\sf \dfrac{2 + {cot}^{2} \theta- {cosec}^{2} \theta }{{cosec}^{2} \theta - 1}$

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$\sf \bigstar Using \:trignometric\:identity \longrightarrow \boxed{\sf 1 +{cot}^{2} \theta = {cosec}^{2} \theta }$

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$\implies \sf \dfrac{2 + {cot}^{2} \theta- (1 +{ cot}^{2} \theta )}{1 + {cot}^{2} \theta - 1}$

$\implies \sf \dfrac{2 + {cot}^{2} \theta- (1 -{ cot}^{2} \theta) }{(1 -1)+{cot}^{2} \theta }$

$\implies \sf \dfrac{2 + {\cancel{cot}}^{2} \theta - 1- {\cancel {cot}}^{2} \theta }{{cot}^{2} \theta }$

$\sf \implies \dfrac{2-1}{{cot}^{2} \theta} \: \: \: \: |in \: num.{cot}^{2} \: get \: cancelled |$

$\sf \implies \dfrac{1}{{cot}^{2} \theta}$

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$\sf \bigstar Using \: trignometric \:identity \longrightarrow \boxed{ \sf \dfrac{1}{{cot}\theta } = {tan} \theta }$

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We get LHS

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$= \sf tan^{2} \theta$

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Therefore,LHS =RHS

Hence proved