JEE Main 2022 — Inverse Trigonometric Functions Question with Solution

Given,

$$\alpha =\tan \left(\frac{5\pi }{16}\sin \left(2{\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)\right)$$

We know, $$2{\cos }^{-1}x={\cos }^{-1}(2x^2-1)$$

$$\therefore$$ $$2{\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)={\cos }^1\left(2\times \frac{1}{5}-1\right)={\cos }^{-1}\left(-\frac{3}{5}\right)$$

$$\therefore$$ $$\alpha =\tan \left(\frac{5\pi }{16}\sin ({\cos }^{-1}\left(-\frac{3}{5}\right)\right)$$

$$=\tan \left(\frac{5\pi }{16}\sin (\pi -{\cos }^{-1}\left(\frac{3}{5}\right)\right)$$

$$=\tan \left(\frac{5\pi }{16}\sin ({\cos }^{-1}\left(\frac{3}{5}\right)\right)$$

$$=\tan \left(\frac{5\pi }{16}\sin ({\sin }^{-1}\left(\frac{4}{5}\right)\right)$$

$$=\tan \left(\frac{5\pi }{16}\times \frac{4}{5}\right)$$

$$=\tan \left(\frac{\pi }{4}\right)$$

$$=1$$

$$\therefore$$ $$\alpha =1$$

Also given,

$$\beta =\cos \left({\sin }^{-1}\left(\frac{4}{5}\right)+{\sec }^{-1}\left(\frac{5}{3}\right)\right)$$

$$=\cos \left({\cos }^{-1}\left(\frac{3}{5}\right)+{\cos }^{-1}\left(\frac{3}{5}\right)\right)$$

$$=\cos \left(2{\cos }^{-1}\left(\frac{3}{5}\right)\right)$$

$$=\cos \left({\cos }^{-1}(2{\left(\frac{3}{5}\right)}^2-1)\right)$$

$$=\cos \left({\cos }^{-1}(\frac{18}{25}-1\right)$$

$$=\frac{18}{25}-1$$

$$=\frac{18-25}{25}$$

$$=-\frac{7}{25}$$

$$\therefore$$ $$\beta =-\frac{7}{25}$$

$$\therefore$$ The quadratic equation with roots $$\alpha$$ and $$\beta$$ is

$$x^2-(\alpha +\beta )x+\alpha \beta =0$$

$$\Rightarrow x^2-\left(1-\frac{7}{25}\right)x+1\times \left(-\frac{7}{25}\right)=0$$

$$\Rightarrow 25x^2-18x-7=0$$