$\frac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }} = $
$\frac{{\sin 3\theta + \sin 5\theta + \sin 7\theta + \sin 9\theta }}{{\cos 3\theta + \cos 5\theta + \cos 7\theta + \cos 9\theta }} = $
- A
$\tan 3\theta $
- B
$\cot 3\theta $
- C
$\tan 6\theta $
- D
$\cot 6\theta $
Similar Questions
यदि $\cos \,(\theta - \alpha ) = a,\,\,\sin \,(\theta - \beta ) = b,\,\,$ हो, तब ${\cos ^2}(\alpha - \beta ) + 2ab\,\sin \,(\alpha - \beta )$ बराबर है
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$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
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- [IIT 1988]
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