$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $
$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $
- A
$\sec A + \tan A$
- B
$\tan \left( {\frac{\pi }{4} - A} \right)$
- C
$\tan \left( {\frac{\pi }{4} + \frac{A}{2}} \right)$
- D
$\tan \left( {\frac{\pi }{4} - \frac{A}{2}} \right)$
Similar Questions
The value of $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$ is $............$.
The value of $\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}$ is $............$.
- [JEE MAIN 2023]
If $cos A = {3\over 4} , $ then $32\sin \left( {\frac{A}{2}} \right)\sin \left( {\frac{{5A}}{2}} \right) = $
If $cos A = {3\over 4} , $ then $32\sin \left( {\frac{A}{2}} \right)\sin \left( {\frac{{5A}}{2}} \right) = $
If $\tan \alpha = \frac{1}{7},\;\tan \beta = \frac{1}{3},$ then $\cos 2\alpha = $
If $\tan \alpha = \frac{1}{7},\;\tan \beta = \frac{1}{3},$ then $\cos 2\alpha = $
If $\frac{{\cos x}}{a} = \frac{{\cos (x + \theta )}}{b} = \frac{{\cos (x + 2\theta )}}{c} = \frac{{\cos (x + 3\theta )}}{d} \, ,$ then $\left( {\frac{{a + c}}{{b + d}}} \right)$ is equal to :-
If $\frac{{\cos x}}{a} = \frac{{\cos (x + \theta )}}{b} = \frac{{\cos (x + 2\theta )}}{c} = \frac{{\cos (x + 3\theta )}}{d} \, ,$ then $\left( {\frac{{a + c}}{{b + d}}} \right)$ is equal to :-
$\tan 3A - \tan 2A - \tan A = $
$\tan 3A - \tan 2A - \tan A = $