${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8}$ =
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8}$ =
- A
$1$
- B
$2$
- C
$1\frac{1}{8}$
- D
$2\frac{1}{8}$
Similar Questions
${\cos ^2}\,{10^o}\,\, - \,\cos \,\,{10^o}\,\cos \,\,{50^o}\, + \,{\cos ^2}\,{50^o}$ ની કિમત ..... થાય.
${\cos ^2}\,{10^o}\,\, - \,\cos \,\,{10^o}\,\cos \,\,{50^o}\, + \,{\cos ^2}\,{50^o}$ ની કિમત ..... થાય.
- [JEE MAIN 2019]
$\frac{{\sin {{81}^o} + \cos {{81}^o}}}{{\sin {{81}^o} - \cos {{81}^o}}}$=
$\frac{{\sin {{81}^o} + \cos {{81}^o}}}{{\sin {{81}^o} - \cos {{81}^o}}}$=
જો $2\sec 2\alpha = \tan \beta + \cot \beta ,$ તો $\alpha + \beta =. . . .$
જો $2\sec 2\alpha = \tan \beta + \cot \beta ,$ તો $\alpha + \beta =. . . .$
$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ =
$cot 5^o$ -$tan5^o$ -$2$ $tan10^o$ -$4$ $tan 20^o$ -$8$ $cot40^o$ =
$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $
$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $