यदि $2 \sin \left(\theta+\frac{\pi}{3}\right)=\cos \left(\theta-\frac{\pi}{6}\right)$ है,तो $\tan \theta=$

  • A
    $\sqrt{3}$
  • B
    $-\frac{1}{\sqrt{3}}$
  • C
    $\frac{1}{\sqrt{3}}$
  • D
    $-\sqrt{3}$

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मान लीजिए $\theta_1$ और $\theta_2$ इस प्रकार हैं कि $(\theta_1-\theta_2)$,$3^{\text{rd}}$ या $4^{\text{th}}$ चतुर्थांश में स्थित है। यदि $\sin \theta_1+\sin \theta_2=-\frac{21}{65}$ और $\cos \theta_1+\cos \theta_2=-\frac{27}{65}$ है,तो $\cos \left(\frac{\theta_1-\theta_2}{2}\right)=$

$\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = $

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