Mathematics - Trigonometric Equation Question with Solution | TestHub

$4\left(\frac{1-{\tan }^2\theta /2}{1+{\tan }^2\theta /2}\right)-3\left(\frac{2\tan \frac{\theta }{2}}{1+{\tan }^2\frac{\theta }{2}}\right)=1$ let $\tan \frac{\theta }{2}=\mathrm{t}$ $\begin{array}{rl}& \frac{4-4t^2-6t}{1+t^2}=1\\ & 4-4t^2-6t=1+t^2\end{array}$ $\begin{array}{rl}& \Rightarrow 5t^2+6t-3=0\\ & \Rightarrow t=\frac{-6\pm \sqrt{36-4(5)(-3)}}{2(5)}\\ & =\frac{-6\pm \sqrt{96}}{10}\\ & =\frac{-6\pm 4\sqrt{6}}{10}\\ & t=\frac{-3+2\sqrt{6}}{5}\\ & \cos \theta =\frac{1-t^2}{1+t^2}=\frac{1-{\left(\frac{2\sqrt{6}-3}{5}\right)}^2}{1+{\left(\frac{2\sqrt{6}-3}{5}\right)}^2}=\frac{1-\left(\frac{24+9-12\sqrt{6}}{25}\right)}{1+\left(\frac{24+9-12\sqrt{6}}{25}\right)}\\ & =\frac{25-33+12\sqrt{6}}{25+33-12\sqrt{6}}=\frac{12\sqrt{6}-8}{58-12\sqrt{6}}=\frac{6\sqrt{6}-4}{29-6\sqrt{6}}\times \frac{29+6\sqrt{6}}{29+6\sqrt{6}}\\ & =\frac{100+150\sqrt{6}}{625}=\frac{4+6\sqrt{6}}{25}\times \frac{4-6\sqrt{6}}{4-6\sqrt{6}}\\ & =\frac{-200}{25(4-6\sqrt{6})}=\frac{-8}{4-6\sqrt{6}}=\frac{4}{3\sqrt{6}-2}\end{array}$