Mathematics - Trigonometric Equation Question with Solution | TestHub

$\frac{1+\tan \frac{\mathrm{y}}{2}}{1-\tan \frac{\mathrm{y}}{2}}={\left(\frac{1+\tan \frac{\mathrm{x}}{2}}{1-\tan \frac{\mathrm{x}}{2}}\right)}^3$

Square both sides, we get

$\frac{1+\sin \mathrm{y}}{1-\sin \mathrm{y}}=\frac{(1+\sin \mathrm{x}{)}^3}{(1-\sin \mathrm{x}{)}^3}$

Using componendo and dividendo

$\frac{2\sin \mathrm{y}}{2}=\frac{\left(3+{\sin }^2\mathrm{x}\right)}{1+3{\sin }^2\mathrm{x}}\sin \mathrm{x}$