સમીકરણ $\sin x \cos x = 2$ ને

  • A
    એક ઉકેલ છે
  • B
    બે ઉકેલ છે
  • C
    અનંત ઉકેલો છે
  • D
    કોઈ ઉકેલ નથી

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$\frac{\cos 15^{\circ} \cos^2 22\frac{1}{2}^{\circ} - \sin 75^{\circ} \sin^2 52\frac{1}{2}^{\circ}}{\cos^2 15^{\circ} - \cos^2 75^{\circ}} = $

સૌથી નાનો પૂર્ણાંક $n$ શોધો જેથી $\frac{1}{\sin 45^{\circ} \sin 46^{\circ}}+\frac{1}{\sin 47^{\circ} \sin 48^{\circ}}+\ldots+\frac{1}{\sin 133^{\circ} \sin 134^{\circ}}=\frac{1}{\sin \left(n^{\circ}\right)}$ થાય.

$\frac{\sin 1^{\circ}+\sin 2^{\circ}+\ldots+\sin 89^{\circ}}{2(\cos 1^{\circ}+\cos 2^{\circ}+\ldots+\cos 44^{\circ})+1} = $

ધારો કે $\theta, \phi \in [0, 2\pi]$ એવા છે કે $2 \cos \theta(1-\sin \phi) = \sin^2 \theta \left(\tan \frac{\theta}{2} + \cot \frac{\theta}{2}\right) \cos \phi - 1$,$\tan (2\pi - \theta) > 0$ અને $-1 < \sin \theta < -\frac{\sqrt{3}}{2}$. તો $\phi$ કઈ શરતનું પાલન કરી શકતું નથી?

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