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

Question

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

Aicharya · Accepted Answer

$	an 70^{\circ}-	an 20^{\circ}=a \cdot 	an 50^{\circ}$ $\frac{\sin 70^{\circ}}{\cos 70^{\circ}}-\frac{\sin 20^{\circ}}{\cos 20^{\circ}}=\frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ $\Rightarrow \quad \frac{\sin 70^{\circ} \cdot \cos 20^{\circ}-\sin 20^{\circ} \cdot \cos 70^{\circ}}{\cos 70^{\circ} \cos 20^{\circ}}=\frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ $[\because \sin (A-B)=\sin A \cos B-\sin B \cos A]$ $\Rightarrow \quad \frac{\sin \left(70^{\circ}-20^{\circ}ight)}{\cos 70^{\circ} \cdot \cos 20^{\circ}}=\frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ $\frac{2 \sin 50^{\circ}}{2 \cos 70 \cdot \cos 20^{\circ}}=\frac{a \sin 50^{\circ}}{\cos 50^{\circ}}$ $[2 \cos A \cos B=\cos (A+B)+\cos (A-B)]$ $\begin{aligned} & \Rightarrow \quad a\left\{2 \cos 70^{\circ} \cdot \cos 20^{\circ}ight\}=2 \cos 50^{\circ} \ & \Rightarrow \quad a\left\{\cos 90^{\circ}+\cos 50^{\circ}ight\}=2 \cos 50^{\circ} \ & \Rightarrow \quad a\left\{\cos 50^{\circ}ight\}=2 \cos 50^{\circ} \Rightarrow a=2\end{aligned}$

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

Solution