If $\theta $ is an acute angle and $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, then $\tan \theta $ is equal to
If $\theta $ is an acute angle and $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, then $\tan \theta $ is equal to
- A
${x^2} - 1$
- B
$\sqrt {{x^2} - 1} $
- C
$\sqrt {{x^2} + 1} $
- D
${x^2} + 1$
Similar Questions
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
- [IIT 1977]
If $x\, sin \theta = y\, sin \, \left( {\theta \,\, + \,\,\frac{{2\,\pi }}{3}} \right) = z\, sin \, \left( {\theta \,\, + \,\,\frac{{4\,\pi }}{3}} \right)$ then :
If $x\, sin \theta = y\, sin \, \left( {\theta \,\, + \,\,\frac{{2\,\pi }}{3}} \right) = z\, sin \, \left( {\theta \,\, + \,\,\frac{{4\,\pi }}{3}} \right)$ then :
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $ is equal to
- [IIT 1975]
The value of $\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$
The value of $\frac{{3 + \cot {{76}^o}\cot {{16}^o}}}{{\cot {{76}^o} + \cot {{16}^o}}}$
$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $
$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $