જો $\tan \frac{\theta }{2} = t,$ તો $\frac{{1 - {t^2}}}{{1 + {t^2}}} = . . . .$
જો $\tan \frac{\theta }{2} = t,$ તો $\frac{{1 - {t^2}}}{{1 + {t^2}}} = . . . .$
- A
$\cos \theta $
- B
$\sin \theta$
- C
$\sec \theta $
- D
$\cos 2\theta $
Similar Questions
$2\sin A{\cos ^3}A - 2{\sin ^3}A\cos A = $
$2\sin A{\cos ^3}A - 2{\sin ^3}A\cos A = $
જો $\tan x = \frac{b}{a},$ તો $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}} = $
જો $\tan x = \frac{b}{a},$ તો $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}} = $
$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $
$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $
- [IIT 1988]
$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $
$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $
$ \cos ^{3}\left(\frac{\pi}{8}\right) \cdot \cos \left(\frac{3 \pi}{8}\right)+\sin ^{3}\left(\frac{\pi}{8}\right) \cdot \sin \left(\frac{3 \pi}{8}\right)$ ની કિમંત મેળવો.
$ \cos ^{3}\left(\frac{\pi}{8}\right) \cdot \cos \left(\frac{3 \pi}{8}\right)+\sin ^{3}\left(\frac{\pi}{8}\right) \cdot \sin \left(\frac{3 \pi}{8}\right)$ ની કિમંત મેળવો.
- [JEE MAIN 2020]