${\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

    $-1$

  • C

    $0$

  • D

    $2$

Similar Questions

 $(sinx + cosecx)^2 + (cosx + secx)^2 - ( tanx + cotx)^2$ =

$\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right) = $

  • [IIT 1984]

જો $\sin \left( {x + \frac{{4\pi }}{9}} \right) = a;\,$ $\frac{\pi }{9}\, < \,x\, < \,\frac{\pi }{3},$ થાય તો $\cos \left( {x + \frac{{7\pi }}{9}} \right)$ = 

જો $x + y = 3 - cos4\theta$ અને $x - y = 4 \,sin2\theta$ હોય તો 

${\rm{cosec }}A - 2\cot 2A\cos A = $