Question

If $x=\tan A+\sin A$ and $y=\tan A-\sin A$, then prove that $x^2-y^2=4\sqrt{xy}$

Answer 1

$x=tanA+sinAandy=tanA-sinA$

considering left hand side,

$\begin{array}{c}{x}^2-y^2=\left(x-y\right)\left(x+y\right)\\ =\left(\left(\tan A+\sin \right)-\left(\tan A-\sin A\right)\right)\left(\left(\tan A+\sin \right)+\left(\tan A-\sin A\right)\right)\\ =2\tan A\times 2\sin A\\ =4\sin A\tan A\end{array}$

Considering right hand side,

$\begin{array}{c}4\sqrt{xy}=4\sqrt{\left(\tan A+\sin A\right)\left(\tan A-\sin A\right)}\\ =4\sqrt{{\tan }^{2}A-{\sin }^{2}A}\\ =4\sqrt{\frac{{\sin }^{2}A}{{\cos }^{2}A}-{\sin }^{2}A}\\ =4\sqrt{{\sin }^{2}A\left(\frac{1-{\cos }^{2}A}{{\cos }^{2}A}\right)}\\ =4\sin A\sqrt{\frac{{\sin }^{2}A}{{\cos }^{2}A}}\\ =4\sin A\sqrt{{\tan }^{2}A}\\ =4\sin A\tan A\end{array}$

So,

LHS=RHS

Hence proved