Jee Main 2024 29 01 2024 Shift 1 - Question7

Answer: (4)

Solution:

$16 \cos ^{2} \theta+25 \sin ^{2} \theta+40 \sin \theta \cos \theta=1$

$16+9 \sin ^{2} \theta+20 \sin 2 \theta=1$

$16+9\left(\frac{1-\cos 2 \theta}{2}\right)+20 \sin 2 \theta=1$

$\frac{-9}{2} \cos 2 \theta+20 \sin 2 \theta=\frac{-39}{2}$

$-9 \cos 2 \theta+40 \sin 2 \theta=-39$

$-9\left(\frac{1-\tan ^{2} \theta}{1+\tan ^{2} \theta}\right)+40\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)=-39$

$48 \tan ^{2} \theta+80 \tan \theta+30=0$

$24 \tan ^{2} \theta+40 \tan \theta+15=0$

$\tan \theta=\frac{-40 \pm \sqrt{(40)^{2}-15 \times 24 \times 4}}{2 \times 24}$

$\tan \theta=\frac{-40 \pm \sqrt{160}}{2 \times 24}$

$=\frac{-10 \pm \sqrt{10}}{12}$

$\Rightarrow \tan \theta=\frac{\sqrt{10}-10}{12}, \quad \tan \theta=\frac{-\sqrt{10}-10}{12}$

So $\tan \theta=-\frac{\sqrt{10}-10}{12}$ will be rejected as $\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

Option (4) is correct.