Prove that (90^ -A ) (90^ -A )= A 1+ ^2 A - Vidyadip

Question

$
\text { Prove that } \sin \left(90^{\circ}-A\right) \cdot \cos \left(90^{\circ}-A\right)=\frac{\tan A}{1+\tan ^2 A}
$

✓

Answer

$
\begin{aligned}
& \text { LHS }=\sin \left(90^{\circ}-A\right) \cdot \cos \left(90^{\circ}-A\right) \\
& \Rightarrow \cos A \cdot \sin A \\
& \text { RHS }=\frac{\tan A}{1+\tan ^2 A}=\frac{\tan A}{\sec ^2 A}=\frac{\frac{\sin A}{\cos A}}{\frac{1}{\cos ^2 A}} \\
& \Rightarrow \text { RHS }=\frac{\sin A}{\cos A} \cdot \cos ^2 A=\cos A \cdot \sin A
\end{aligned}
$
Thus , $\mathrm{LHS}=\mathrm{RHS}$
$
\Rightarrow \sin \left(90^{\circ}-A\right) \cdot \cos \left(90^{\circ}-A\right)=\frac{\tan A}{1+\tan ^2 A}
$

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