Question

If $90°<A<180°$and $\sin A=4/5$, then $\tan A/2$ is equal to

Answer 1

Finding the $\mathit{t}\mathit{a}\mathit{n}\mathit{A}\mathbf{/}\mathbf{2}$ :

$$\begin{gathered} \sin A=4/5,90°<A<180° \\ \sin A=4/5, \\ \sin ce,\cos A=\pm \sqrt{1-{\sin }^{2}A} \\ So,\cos A=\pm \sqrt{1-16/25} \\ \cos A=\pm 3/5, \\ \cos A=-3/5(\because 90°<A<180°) \\ Also,\cos A=2{\cos }^{2}A/2-1=-3/5 \\ 2{\cos }^{2}A/2=2/5 \\ \cos A/2=\pm \sqrt{2/10} \\ \cos A/2=\sqrt{2/10}(\because 45°<A/2<90°) \\ \sin A/2=\pm \sqrt{1-(2/10)} \\ \sin A/2=\sqrt{8/10}(\because 45°<A/2<90°) \\ \tan A/2=(\sin A/2)/\cos A/2 \\ \tan A/2=\left(\sqrt{8/10}\right)/\sqrt{2/10} \\ \tan A/2=\sqrt{4}=2 \end{gathered}$$

Hence the value of $\mathit{t}\mathit{a}\mathit{n}\mathit{A}\mathbf{/}\mathbf{2}\mathbf{=}\mathbf{2}\mathbf{.}$