$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (when $x$ lies in $II^{nd}$ quadrant)
$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (when $x$ lies in $II^{nd}$ quadrant)
- A
$\sin \frac{x}{2}$
- B
$\tan \frac{x}{2}$
- C
$\sec \frac{x}{2}$
- D
${\rm{cosec}}\frac{x}{2}$
Similar Questions
Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
Prove that: $\cos 6 x=32 x \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$
Prove that: $\cos 6 x=32 x \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$
If $\tan \frac{\theta }{2} = t,$then $\frac{{1 - {t^2}}}{{1 + {t^2}}}$is equal to
If $\tan \frac{\theta }{2} = t,$then $\frac{{1 - {t^2}}}{{1 + {t^2}}}$is equal to
If $A, B, C$ are angles of a triangle, then $\sin 2A + \sin 2B - \sin 2C$ is equal to
If $A, B, C$ are angles of a triangle, then $\sin 2A + \sin 2B - \sin 2C$ is equal to
If $\sin \theta = \frac{1}{2}\left( {\sqrt {\frac{x}{y}\,} + \,\sqrt {\frac{y}{x}} } \right)\,,\,\left( {x,y \in R\, - \{ 0\} } \right)$. Then
If $\sin \theta = \frac{1}{2}\left( {\sqrt {\frac{x}{y}\,} + \,\sqrt {\frac{y}{x}} } \right)\,,\,\left( {x,y \in R\, - \{ 0\} } \right)$. Then