tan 81^{circ}-tan 63^{circ}-tan 27^{circ}+tan | vedclass

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Question

(D) આપણી પાસે પદાવલિ છે: $	an 81^{\circ} + 	an 9^{\circ} - (	an 63^{\circ} + 	an 27^{\circ})$નિત્યસમ $	an A + 	an B = \frac{\sin(A+B)}{\cos A \c

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) આપણી પાસે પદાવલિ છે: $	an 81^{\circ} + 	an 9^{\circ} - (	an 63^{\circ} + 	an 27^{\circ})$નિત્યસમ $	an A + 	an B = \frac{\sin(A+B)}{\cos A \cos B}$ નો ઉપયોગ કરતા:$	an 81^{\circ} + 	an 9^{\circ} = \frac{\sin(81^{\circ}+9^{\circ})}{\cos 81^{\circ} \cos 9^{\circ}} = \frac{\sin 90^{\circ}}{\cos 81^{\circ} \cos 9^{\circ}} = \frac{1}{\sin 9^{\circ} \cos 9^{\circ}} = \frac{2}{\sin 18^{\circ}}$તે જ રીતે,$	an 63^{\circ} + 	an 27^{\circ} = \frac{\sin(63^{\circ}+27^{\circ})}{\cos 63^{\circ} \cos 27^{\circ}} = \frac{\sin 90^{\circ}}{\cos 63^{\circ} \cos 27^{\circ}} = \frac{1}{\sin 27^{\circ} \cos 27^{\circ}} = \frac{2}{\sin 54^{\circ}}$હવે,પદાવલિ બને છે $\frac{2}{\sin 18^{\circ}} - \frac{2}{\sin 54^{\circ}} = 2 \left( \frac{\sin 54^{\circ} - \sin 18^{\circ}}{\sin 18^{\circ} \sin 54^{\circ}} ight)$$\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$ નો ઉપયોગ કરતા:$\sin 54^{\circ} - \sin 18^{\circ} = 2 \cos 36^{\circ} \sin 18^{\circ}$તેથી,$2 \left( \frac{2 \cos 36^{\circ} \sin 18^{\circ}}{\sin 18^{\circ} \cos 36^{\circ}} ight) = 2 	imes 2 = 4$

$\tan 81^{\circ}-\tan 63^{\circ}-\tan 27^{\circ}+\tan 9^{\circ}$ ની કિંમત શોધો.

Similar Questions

પદાવલિના અવયવ પાડો: $\sec ^2 x+5 \tan x+5$

ધારો કે $S = \left\{ \theta \in \left( 0, \frac{\pi}{2} \right) : \sum_{m=1}^{9} \sec \left( \theta + (m-1) \frac{\pi}{6} \right) \sec \left( \theta + \frac{m \pi}{6} \right) = -\frac{8}{\sqrt{3}} \right\}$. તો:

$\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}} = $

જો $\alpha$ એ $25\cos^2\theta + 5\cos\theta - 12 = 0$ નું બીજ હોય અને $\pi/2 < \alpha < \pi$ હોય,તો $\sin 2\alpha$ ની કિંમત શોધો:

$\cos^2 76^{\circ} + \cos^2 16^{\circ} - \cos 76^{\circ} \cos 16^{\circ}$ ની કિંમત શોધો.

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