STD 12 - 2. inverse trigonometric function — Mathematics STD 12 Science — Question

Let $\tan ^{-1} 2=\theta$

$2=\tan \theta$

$x=\sin 2 \theta$

$x=2 \sin \theta \cos \theta$

$x=2 \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}=\frac{4}{5}$

$\because \quad 1-x=1 / 5$

and $ \quad \because y^{2}=\frac{1}{5}$

$\tan ^{-1} 4 / 3=\alpha$

$4 / 3=\tan \alpha$

$y=\sin \frac{\alpha}{2}$

$y=\sqrt{\frac{1-\cos \alpha}{2}}=\sqrt{\frac{1-3 / 5}{2}}$

$y=\frac{1}{\sqrt{5}} \quad \therefore \quad y^{2}=1-x$