Question

Please help me prove both the sides equal ​

Answer 1

Answer:

The Answer Is in attachment Image.

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Answer 2

$\huge{ \underline{ \mathfrak{ \green{ \: Answer}}}}$

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$\boxed{\red{\bold{Solving\: LHS:}}}$

$sin(90^o - \theta) × cos( 90^o - \theta)$

$cos \theta × sin \theta$

$\mathcal{\underline{\blue{ Because \:sin(90^o - \theta) = cos \theta}}}$

$\mathcal{\underline{\blue{ And \:cos(90^o -\theta) = sin \theta}}}$

$\boxed{\red{\bold{Solving\: RHS:}}}$

$\displaystyle{\frac{tan \theta}{1 +tan^2 \theta}}$

$\displaystyle{\frac{tan \theta}{sec^2 \theta}}$

$\mathcal{\underline{\blue{Because \: 1 + tan^2 \theta = sec^2 \theta}}}$

$\displaystyle{\frac{sin \theta × cos^2 \theta}{cos \theta}}$

$\mathcal{\underline{\blue{Because \: sec^2 \theta}}}= \mathcal{\blue{\displaystyle{\frac{1}{cos^2 \theta}}}}$

$\boxed{\red{\bold{Canceling \; cos \theta:}}}$

$sin \theta × cos \theta$

$\bold{\blue{\mathcal{Hence\: LHS=RHS}}}$

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