Question

If $\alpha =ta{n}^{-1}\left(\frac{4x-4x^3}{1-6x^2+x^4}\right),\beta =2si{n}^{-1}\left(\frac{2x}{1+x^2}\right)$ and $tan\frac{\pi }{8}=k$, then

• A. $\alpha +\beta =\pi$ for x $ϵ[1,\frac{1}{k})$
• B. $\alpha =\beta$ for x $ϵ(-k,k)$
• C. $\alpha +\beta =-\pi$ for x $ϵ[1,\frac{1}{k})$
• D. $\alpha +\beta =0$ for x $ϵ(-k,k)$

Answer 1

Answer: (A), (B)

The correct options are
A

$\alpha +\beta =\pi$ for x $ϵ[1,\frac{1}{k})$

B

$\alpha =\beta$ for x $ϵ(-k,k)$

Put x = tan $\theta$

$\Rightarrow \alpha =ta{n}^{-1}\left(tan4\theta \right)$

$=4\theta -\pi$ for x $ϵ[1,\frac{1}{k})$

$=4\theta$ for x $ϵ(-k,k)$

Also $\beta =2(\pi -2\theta )$ for x $ϵ[1,\frac{1}{k})$

$=4\theta$ for x $ϵ(-k,k)$