$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $
$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $
- A
$\tan A$
- B
$\tan 2A$
- C
$\cot A$
- D
$\cot 2A$
Similar Questions
If $k = \sin \frac{\pi }{{18}}\,.\,\sin \frac{{5\pi }}{{18}}\,.\,\sin \frac{{7\pi }}{{18}},$ then the numerical value of $k$ is
If $k = \sin \frac{\pi }{{18}}\,.\,\sin \frac{{5\pi }}{{18}}\,.\,\sin \frac{{7\pi }}{{18}},$ then the numerical value of $k$ is
- [IIT 1993]
સાબિત કરો કે : $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
સાબિત કરો કે : $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $
$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $
સમીકરણ $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ =
સમીકરણ $\frac{{{{\tan }^2}20^\circ - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ =
$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $
$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $