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Answer 1

tan^2o - sin^2o = tan^2o.sin^2o

sin^2o/cos^2o - sin^2o/1 = tan^2o.sin^2o

sin^2o - cos^2o sin^2o/ cos^2o = tan^2o.sin^2o

sin^2o(1-cos^2o)/cos^2o = tan^2o.sin^2o

sin^2o×(sin^2o)/cos^2=tan^2o.sin^2o

sin^2o . tan^2o = tan^2o.sin^2o

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Answer 2

$\large\underline{\underline{\frak{\qquad Solution:\qquad}}}$

$:\implies \sf LHS = \tan^2 ( \theta) - \sin^{2}(\theta)$

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

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

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

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

$:\implies \sf \sin^{2}(\theta) .\tan^{2} (\theta)$

$:\implies \underline{ \boxed{ \sf \tan^{2} (\theta). \sin^{2}(\theta) =R.H.S }}$

$:\implies \underline{ \boxed{ \sf L.H.S = R.H.S }}$

$:\implies \underline{ \boxed{ \sf Hence \: Proved. }}$

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$\large\underline{\underline{\frak{\qquad Formuals \: related \: to \: it:\qquad}}}$

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

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