We have, tanx = $-\frac{4}{3}$ , x in quadrant II since , x in quadrant II ⇒ $\frac{\pi}{2}$ < x < π , ⇒ $\frac{\pi}{4}$ < $\frac{x}{2}$ < $\frac{\pi}{2}$ ⇒ $\frac{x}{2}$ lies in 1st quadrant ⇒ sin $\frac{x}{2}$ > 0 , cos $\frac{x}{2}$ > 0 , tan $\frac{x}{2}$ > 0 Also 1 + $\tan^2 x$ = $\sec^2 x$ ⇒ $\sec^2 x$ ⇒ $\sec^2 x$ = 1 +$\frac{16}{9}$ = $\frac{25}{9}$ ⇒ sec x = ± $\frac{5}{3}$ ⇒ cos x = - 3/5 Since $\frac{\pi}{2}$ < x < π , ∴ cos x is - ve Now cos $\frac{x}{2}$ = ± $\sqrt{\frac{1+\cos x}{2}}$ = $\sqrt{\frac{1-\frac{3}{5}}{2}}$ Since cos $\frac{x}{2}$ is + ve = $\sqrt{\frac{2}{5} \times \frac{1}{2}}$ = $\sqrt{\frac{1}{5}}$ = $\frac{1}{\sqrt{5}}$ sin $\frac{x}{2}$ = ± $\sqrt{\frac{1-\cos x}{2}}$ = ± $\sqrt{\frac{1+\frac{3}{5}}{2}}$ = $\sqrt{\frac{8}{5} \times \frac{1}{2}}$ = $\frac{2}{\sqrt{5}}$ Since sin $\frac{x}{2}$ > 0 tan $\frac{x}{2}$ = $\frac{\sin x}{2} \cdot \frac{\cos x}{2}$ = $\frac{2}{\sqrt{5}} \times \frac{\sqrt{5}}{1}$ = 2 Hence sin $\frac{x}{2}$ = $\frac{2\sqrt{5}}{2}$ , cos $\frac{x}{2}$ = $\frac{\sqrt{5}}{2}$ , tan $\frac{x}{2}$ = 2