Question

If $\tan \theta +\sin \theta =a$ and $\tan \theta -\sin \theta =b$ what is the value $(a^2-b^2)÷\sqrt{ab}$.

Answer 1

$a=\tan \theta +\sin \theta$
$b=\tan \theta -\sin \theta$
$\therefore a+b=2\tan \theta$ and $a-b=2\sin \theta$
$a^2-b^2=(a+b)(a-b)=4\sin \theta \tan \theta$
Again $\sqrt{ab}=\sqrt{(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )}$
$=\sqrt{{\tan }^2\theta -{\sin }^2\theta }$
$=\sqrt{\frac{{\sin }^2\theta }{{\cos }^2\theta }}-{\sin }^2\theta =\sqrt{{\sin }^2\theta \left({\sec }^2\theta \right)-{\sin }^2\theta }$
$=\sqrt{{\sin }^2\theta ({\sec }^2\theta -1)}=\sqrt{{\sin }^2\theta .{\tan }^2\theta }$
$=\sin \theta \tan \theta$
$=\frac{a^2-b^2}{\sqrt{ab}}=\frac{4\sin \theta \tan \theta }{\sin \theta \tan \theta }=4$