Determine the value of a in tan 70^{circ} - t | vedclass

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(D) Given: $	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) Given: $	an 70^{\circ} - 	an 20^{\circ} = a \cdot 	an 50^{\circ}$ $\Rightarrow \frac{\sin 70^{\circ}}{\cos 70^{\circ}} - \frac{\sin 20^{\circ}}{\cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ $\Rightarrow \frac{\sin 70^{\circ} \cos 20^{\circ} - \sin 20^{\circ} \cos 70^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$: $\Rightarrow \frac{\sin(70^{\circ} - 20^{\circ})}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ $\Rightarrow \frac{\sin 50^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ Since $\sin 50^{\circ} 
eq 0$,we can divide both sides by $\sin 50^{\circ}$: $\Rightarrow \frac{1}{\cos 70^{\circ} \cos 20^{\circ}} = \frac{a}{\cos 50^{\circ}}$ $\Rightarrow a = \frac{\cos 50^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}}$ Multiply numerator and denominator by $2$: $a = \frac{2 \cos 50^{\circ}}{2 \cos 70^{\circ} \cos 20^{\circ}}$ Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$: $a = \frac{2 \cos 50^{\circ}}{\cos(70^{\circ} + 20^{\circ}) + \cos(70^{\circ} - 20^{\circ})}$ $a = \frac{2 \cos 50^{\circ}}{\cos 90^{\circ} + \cos 50^{\circ}}$ Since $\cos 90^{\circ} = 0$: $a = \frac{2 \cos 50^{\circ}}{0 + \cos 50^{\circ}} = \frac{2 \cos 50^{\circ}}{\cos 50^{\circ}} = 2$

Determine the value of $a$ in $\tan 70^{\circ} - \tan 20^{\circ} = a \cdot \tan 50^{\circ} ?$

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