sin 50^circ - sin 70^circ + sin 10^circ = | vedclass

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(B) $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$ સૂત્રનો ઉપયોગ કરતા:$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ$$= 2 \cos \frac{50^\

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(B) $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$ સૂત્રનો ઉપયોગ કરતા:$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ$$= 2 \cos \frac{50^\circ + 70^\circ}{2} \sin \frac{50^\circ - 70^\circ}{2} + \sin 10^\circ$$= 2 \cos 60^\circ \sin(-10^\circ) + \sin 10^\circ$$= -2 \cos 60^\circ \sin 10^\circ + \sin 10^\circ$$= -2 	imes \frac{1}{2} 	imes \sin 10^\circ + \sin 10^\circ$$= -\sin 10^\circ + \sin 10^\circ = 0.$

$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ = $

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$\cos 1^\circ \cdot \cos 2^\circ \cdot \cos 3^\circ \cdot \dots \cdot \cos 179^\circ = $

જો $\sin x = -\frac{24}{25}$ હોય,તો $\tan x$ ની કિંમત શોધો.

પદાવલિ $\frac{\tan(x - \frac{\pi}{2}) \cdot \cos(\frac{3\pi}{2} + x) - \sin^3(\frac{7\pi}{2} - x)}{\cos(x - \frac{\pi}{2}) \cdot \tan(\frac{3\pi}{2} + x)}$ નું સાદું રૂપ આપતા શું મળે?

પદાવલિનું સાદું રૂપ આપો: $\frac{\cos x}{1+\sin x} + \tan x$

કિંમત શોધો: $\tan A + \cot (180^\circ + A) + \cot (90^\circ + A) + \cot (360^\circ - A)$

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