$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $
$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $
- [IIT 1974]
- A
$1$
- B
$2$
- C
$3$
- D
$\sqrt 3 /2$
Similar Questions
$\cos \alpha .\sin (\beta - \gamma ) + \cos \beta .\sin (\gamma - \alpha ) + \cos \gamma .\sin (\alpha - \beta ) = $
$\cos \alpha .\sin (\beta - \gamma ) + \cos \beta .\sin (\gamma - \alpha ) + \cos \gamma .\sin (\alpha - \beta ) = $
निम्नलिखित को सिद्ध कीजिए
$\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$
निम्नलिखित को सिद्ध कीजिए
$\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$
यदि $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
तथा $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ हो, तब
यदि $\tan x = \frac{{2b}}{{a - c}}(a \ne c),$
$y = a\,{\cos ^2}x + 2b\,\sin x\cos x + c\,{\sin ^2}x$
तथा $z = a{\sin ^2}x - 2b\sin x\cos x + c{\cos ^2}x,$ हो, तब
${\sin ^4}\frac{\pi }{4} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
${\sin ^4}\frac{\pi }{4} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $
$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $