Question

The trigonometric equation $\frac{1-\sin \theta }{1+\sin \theta }$ is equal to
[Given, ${\sin }^2\theta +{\cos }^2\theta =1$]

• A. $(\sec \theta +\tan \theta {)}^2$
• B. $(\cot \theta -\tan \theta {)}^2$
• C. $(\sec \theta -\tan \theta {)}^2$
• D. $(\text{cosec}\theta -\tan \theta {)}^2$

Answer 1

The correct option is C $(\sec \theta -\tan \theta {)}^2$

Multiplying both numerator and denominator by $(1-\sin \theta )$.

$\frac{1-\sin \theta }{1+\sin \theta }=\frac{(1-\sin \theta )\times (1-\sin \theta )}{(1+\sin \theta )\times (1-\sin \theta )}$ $=\frac{(1-\sin \theta {)}^2}{1-{\sin }^2\theta }$
As we have
${\sin }^2\theta +{\cos }^2\theta =1\Rightarrow 1-{\sin }^2\theta ={\cos }^2\theta$
$\therefore \frac{(1-\sin \theta )\times (1-\sin \theta )}{(1+\sin \theta )\times (1-\sin \theta )}=\frac{(1-\sin \theta {)}^2}{{\cos }^2\theta }$$=(\frac{1-\sin \theta }{\cos \theta }{)}^2$$=(\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta }{)}^2$$=(\sec \theta -\tan \theta {)}^2$