tan 70^{circ} - tan 20^{circ} = a cdot tan 50 | vedclass

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Question

(D) આપેલ છે: $	an 70^{\circ} - 	an 20^{\circ} = a \cdot 	an 50^{\circ}$ $\Rightarrow \frac{\sin 70^{\circ}}{\cos 70^{\circ}} - \frac{\sin 20^{\circ

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(D) આપેલ છે: $	an 70^{\circ} - 	an 20^{\circ} = a \cdot 	an 50^{\circ}$ $\Rightarrow \frac{\sin 70^{\circ}}{\cos 70^{\circ}} - \frac{\sin 20^{\circ}}{\cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ $\sin(A-B) = \sin A \cos B - \cos A \sin B$ નો ઉપયોગ કરતા: $\Rightarrow \frac{\sin(70^{\circ} - 20^{\circ})}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ $\Rightarrow \frac{\sin 50^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ $\sin 50^{\circ} 
eq 0$ હોવાથી,બંને બાજુ $\sin 50^{\circ}$ વડે ભાગતા: $\Rightarrow a = \frac{\cos 50^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}}$ અંશ અને છેદને $2$ વડે ગુણતા: $a = \frac{2 \cos 50^{\circ}}{2 \cos 70^{\circ} \cos 20^{\circ}}$ $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ નો ઉપયોગ કરતા: $a = \frac{2 \cos 50^{\circ}}{\cos 90^{\circ} + \cos 50^{\circ}} = \frac{2 \cos 50^{\circ}}{0 + \cos 50^{\circ}} = 2$

$\tan 70^{\circ} - \tan 20^{\circ} = a \cdot \tan 50^{\circ}$ માં $a$ ની કિંમત શોધો.

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