$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $
$\frac{{\sqrt 2 - \sin \alpha - \cos \alpha }}{{\sin \alpha - \cos \alpha }} = $
- A
$\sec \left( {\frac{\alpha }{2} - \frac{\pi }{8}} \right)$
- B
$\cos \left( {\frac{\pi }{8} - \frac{\alpha }{2}} \right)$
- C
$\tan \left( {\frac{\alpha }{2} - \frac{\pi }{8}} \right)$
- D
$\cot \left( {\frac{\alpha }{2} - \frac{\pi }{2}} \right)$
Similar Questions
$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}$ बराबर है
$96 \cos \frac{\pi}{33} \cos \frac{2 \pi}{33} \cos \frac{4 \pi}{33} \cos \frac{8 \pi}{33} \cos \frac{16 \pi}{33}$ बराबर है
- [JEE MAIN 2023]
यदि $\sin 6\theta = 32{\cos ^5}\theta \sin \theta - 32{\cos ^3}\theta \sin \theta + 3x,$ तब $x = $
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$\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 ) = $
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यदि $\frac{{2\sin \alpha }}{{\{ 1 + \cos \alpha + \sin \alpha \} }} = y,$ हो, तो $\frac{{\{ 1 - \cos \alpha + \sin \alpha \} }}{{1 + \sin \alpha }} $ बराबर है
यदि $\frac{{2\sin \alpha }}{{\{ 1 + \cos \alpha + \sin \alpha \} }} = y,$ हो, तो $\frac{{\{ 1 - \cos \alpha + \sin \alpha \} }}{{1 + \sin \alpha }} $ बराबर है