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}}} = $
- A
$\frac{{2\sin x}}{{\sqrt {\sin 2x} }}$
- B
$\frac{{2\cos x}}{{\sqrt {\cos 2x} }}$
- C
$\frac{{2\cos x}}{{\sqrt {\sin 2x} }}$
- D
$\frac{{2\sin x}}{{\sqrt {\cos 2x} }}$
Similar Questions
${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ =
$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ =
સાબિત કરો કે, $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$
સાબિત કરો કે, $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$
જો $\tan A = \frac{1}{2},$ તો $\tan 3A = $
જો $\tan A = \frac{1}{2},$ તો $\tan 3A = $
$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ =
$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ =