Question

${\displaystyle \sqrt{\frac{1-\sin \theta }{1+\sin \theta }}}$ = secθ - tanθ

Answer 1

डावी बाजू = ${\displaystyle \sqrt{\frac{1-\sin \theta }{1+\sin \theta }}}$ 

= ${\displaystyle \sqrt{\frac{1-\sin \theta }{1+\sin \theta }\times \frac{1-\sin \theta }{1-\sin \theta }}}$ .....[छेदाचे परिमेयकरण करून]

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

= ${\displaystyle \sqrt{\frac{{(1-\sin \theta )}^2}{{\cos }^2\theta }}}$ ...${\displaystyle \left[\begin{array}{c}\because {\sin }^2\theta +{\cos }^2\theta =1\\ \therefore 1-{\sin }^2\theta ={\cos }^2\theta \end{array}\right]}$

= ${\displaystyle \frac{1-\sin \theta }{\cos \theta }=\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }}$

= sec θ - tan θ

= उजवी बाजू

∴ ${\displaystyle \sqrt{\frac{1-\sin \theta }{1+\sin \theta }}}$ = sec θ - tan θ