Question

If $\tan \theta +\sin \theta =a$ and $\tan \theta -\sin \theta =b$, then prove that $a^2-b^2=4\sqrt{ab}$

Answer 1

As $\tan \theta +\sin \theta =a$ and $\tan \theta -\sin \theta =b$, then, simplifying the LHS of $a^2-b^2=4\sqrt{ab}$.

$a^2-b^2=\left(a-b\right)\left(a+b\right)$ $=\left(\tan \theta +\sin \theta -\tan \theta +\sin \theta \right)\left(\tan \theta +\sin \theta +\tan \theta -\sin \theta \right)$

$=\left(2\sin \theta \right)\left(2\tan \theta \right)$

$=4\tan \theta \sin \theta$

Now simplifying the RHS of $a^2-b^2=4\sqrt{ab}$.

$4\sqrt{ab}=4\sqrt{\left(\tan \theta +\sin \theta \right)\left(\tan \theta -\sin \theta \right)}$

$=4\sqrt{{\tan }^2\theta -{\sin }^2\theta }$

$=4\sqrt{\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta }$

$=4\sqrt{\frac{{\sin }^2\theta -{\sin }^2\theta {\cos }^2\theta }{{\cos }^2\theta }}$

$=4\sqrt{\frac{{\sin }^2\theta \left(1-{\cos }^2\theta \right)}{{\cos }^2\theta }}$

$=4\sqrt{\frac{{\sin }^2\theta }{{\cos }^2\theta }\times {\sin }^2\theta }$

$=4\sqrt{{\tan }^2\theta {\sin }^2\theta }$

$=4\tan \theta \sin \theta$

This shows that $LHS=RHS$.

Hence proved.