${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
${(\cos \alpha + \cos \beta )^2} + {(\sin \alpha + \sin \beta )^2} = $
- A
$4{\cos ^2}\frac{{\alpha - \beta }}{2}$
- B
$4{\sin ^2}\frac{{\alpha - \beta }}{2}$
- C
$4{\cos ^2}\frac{{\alpha + \beta }}{2}$
- D
$4{\sin ^2}\frac{{\alpha + \beta }}{2}$
Similar Questions
यदि $\alpha + \beta = \frac{\pi }{2}$ तथा $\beta + \gamma = \alpha ,$ तब $\tan \,\alpha $ =
यदि $\alpha + \beta = \frac{\pi }{2}$ तथा $\beta + \gamma = \alpha ,$ तब $\tan \,\alpha $ =
- [IIT 2001]
निम्नलिखित को सिद्ध कीजिए
$\cos 6 x=32 x \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$
निम्नलिखित को सिद्ध कीजिए
$\cos 6 x=32 x \cos ^{6} x-48 \cos ^{4} x+18 \cos ^{2} x-1$
यदि $\sin A = n\sin B,$ तो $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $
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$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $
$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $
$2\sin A{\cos ^3}A - 2{\sin ^3}A\cos A = $
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