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सिद्ध कीजिए कि: $\cot ^{2} \frac{\pi}{6} + \csc \frac{5 \pi}{6} + 3 \tan ^{2} \frac{\pi}{6} = 6$

$\frac{\sin 70^\circ + \cos 40^\circ}{\cos 70^\circ + \sin 40^\circ} = $

$\frac{1}{1+\sin \theta}+\frac{1}{1-\sin \theta} = $

$1+\cot ^2 30^{\circ}-\sec ^2 45^{\circ}=$

यदि $\theta$ तीसरे चतुर्थांश में स्थित है और $\cos \theta = -\frac{3}{5}$ है,तो $\tan \theta$ का मान ज्ञात कीजिए।

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