JEE Main 2024 — 5 April, Shift 1 — Question 15

$4\left(\frac{1-\tan ^{2} \theta / 2}{1+\tan ^{2} \theta / 2}\right)-3\left(\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)=1$

let $\tan \frac{\theta}{2}=\mathrm{t}$

$\frac{4-4 t^{2}-6 t}{1+t^{2}}=1$

$4-4 t^{2}-6 t=1+t^{2}$ $\Rightarrow 5 \mathrm{t}^{2}+6 \mathrm{t}-3=0$

$\Rightarrow \mathrm{t}=\frac{-6 \pm \sqrt{36-4(5)(-3)}}{2(5)}$

$=\frac{-6 \pm \sqrt{96}}{10}$

$=\frac{-6 \pm 4 \sqrt{6}}{10}$

$t=\frac{-3+2 \sqrt{6}}{5}$

$\cos \theta=\frac{1-\mathrm{t}^{2}}{1+\mathrm{t}^{2}}=\frac{1-\left(\frac{2 \sqrt{6}-3}{5}\right)^{2}}{1+\left(\frac{2 \sqrt{6}-3}{5}\right)^{2}}=\frac{1-\left(\frac{24+9-12 \sqrt{6}}{25}\right)}{1+\left(\frac{24+9-12 \sqrt{6}}{25}\right)}$

$=\frac{25-33+12 \sqrt{6}}{25+33-12 \sqrt{6}}=\frac{12 \sqrt{6}-8}{58-12 \sqrt{6}}=\frac{6 \sqrt{6}-4}{29-6 \sqrt{6}} \times \frac{29+6 \sqrt{6}}{29+6 \sqrt{6}}$

$=\frac{100+150 \sqrt{6}}{625}=\frac{4+6 \sqrt{6}}{25} \times \frac{4-6 \sqrt{6}}{4-6 \sqrt{6}}$ $=\frac{-200}{25(4-6 \sqrt{6})}=\frac{-8}{4-6 \sqrt{6}}=\frac{4}{3 \sqrt{6}-2}$