Question

Prove that \(\left(\sin 60^{\circ}+\sin 30^{\circ}\right) /\left(\sin 60^{\circ}-\sin 30^{\circ}\right)=\left(\tan 60^{\circ}+\tan 45^{\circ}\right) /\left(\tan 60^{\circ}-\tan 45^{\circ}\right)\)

Answer 1

$$
\begin{aligned}
&\text { LHS }=\left(\sin 60^{\circ}+\sin 30^{\circ}\right) /\left(\sin 60^{\circ}-\sin 30^{\circ}\right) \\
&=\frac{\frac{\sqrt{3}}{2}+\frac{1}{2}}{\frac{\sqrt{3}}{2}-\frac{1}{2}}=\frac{\frac{\sqrt{3}+1}{2}}{\frac{\sqrt{3}-1}{2}} \\
&\left.=\frac{\sqrt{3}+1}{\sqrt{3}-1}\left[\because \sqrt{3}=\tan 60^{\circ}\right] \quad 1=\tan 45^{\circ}\right] \\
&=\frac{\tan 60^{\circ}+\tan 45^{\circ}}{\tan 60^{\circ}-\tan 45^{\circ}}
\end{aligned}
$$
Thus, L.H.S. = R.H.S.