Question

Prove that $\frac{\sin {60}^o+\sin {30}^o}{\sin {60}^o-\sin {30}^o}=\frac{\tan {60}^o+\tan {45}^o}{\tan {60}^o-\tan {45}^o}$

Answer 1

$\sin 60°=\frac{\sqrt{3}}{2};\sin 30°=\frac{1}{2};\tan 60°=\sqrt{3};\tan 45°=1$

$\frac{\sin 60°+\sin 30°}{\sin 60°-\sin 30°}=\frac{\frac{\sqrt{3}}{2}-\frac{1}{2}}{\frac{\sqrt{3}}{2}-\frac{1}{2}}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$\frac{\tan 60°+\tan 45°}{\tan 60°-\tan 45°}=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\sin 60°+\sin 30°}{\sin 60°-\sin 30°}$ proved