Question

If $\alpha =\tan \left(\frac{1}{3}{\tan }^{-1}\frac{4}{3}\right)$ and $\beta =\sin \left(\frac{1}{2}{\sin }^{-1}\frac{4\sqrt{2}}{9}\right)$, then the value of $\left|\left({\alpha }^3-4{\alpha }^2-3\alpha +\beta \right)\right|$ is equal to

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

Let ${\tan }^{-1}\left(\frac{4}{3}\right)=3\theta ;\alpha =\tan \theta \Rightarrow \tan 3\theta =\frac{4}{3}$
$\Rightarrow \left({\alpha }^3-4{\alpha }^2-3\alpha \right)=\frac{-4}{3}$....(1)
Also, put ${\sin }^{-1}\left(\frac{4\sqrt{2}}{9}\right)=2\varphi ;\beta =\sin \varphi \Rightarrow \sin 2\varphi =\frac{4\sqrt{2}}{9}\Rightarrow \beta =\frac{1}{3}$....(2)
$\therefore \left|{\alpha }^3-4{\alpha }^2-3\alpha +\beta \right|=\left|\frac{-4}{3}+\frac{1}{3}\right|=1$