Question

If $\tan \theta +\sin \theta =a$ and $\tan \theta -\sin \theta =\beta$ show that $a^2-{\beta }^2=4\sqrt{a\beta }$

Answer 1

Given that,

$\tan \theta +\sin \theta =a$

$\tan \theta -\sin \theta =\beta$

$a\beta =(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )$

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

$={\sin }^2\theta (\frac{1}{{\cos }^2\theta }-1)$

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

$={\sin }^2\theta \left(\frac{{\sin }^2\theta }{{\cos }^2\theta }\right)$

$={\sin }^2\theta {\tan }^2\theta$............(i)

$a^2-{\beta }^2=(\tan \theta +\sin \theta {)}^2-(\tan \theta -\sin \theta {)}^2$

$=4\tan \theta \sin \theta$ $[\because (a+b{)}^2-(a-b{)}^2=4ab]$

$=4\sqrt{a\beta }$ [From (i)]

Hence prove.