Question

If $2{\sin }^{-1}\left(\frac{9-4x^2}{9+4x^2}\right)+3{\cos }^{-1}\left(\frac{12x}{9+4x^2}\right)+{\tan }^{-1}\left(\frac{12x}{9-4x^2}\right)=\lambda \frac{\pi }{2}$ for $x=-\frac{17}{13}$, than $\lambda$ is

Answer 1

$2{\sin }^{-1}\left(\frac{9-4x^2}{9+4x^2}\right)+3{\cos }^{-1}\left(\frac{12x}{9+4x^2}\right)+{\tan }^{-1}\left(\frac{12x}{9-4x^2}\right)=\lambda \frac{\pi }{2}$
Putting $\frac{2x}{3}=\tan \theta$ we have
$2{\sin }^{-1}\left(\cos 2\theta \right)+3{\cos }^{-1}\left(\sin 2\theta \right)+{\tan }^{-1}\left(\tan 2\theta \right)=\frac{\lambda \pi }{2}\Rightarrow 2(\frac{\pi }{2}-{\cos }^{-1}\cos 2\theta )+3(\frac{\pi }{2}-{\sin }^{-1}\sin 2\theta )+{\tan }^{-1}\tan 2\theta =\frac{\lambda \pi }{2}$
For $x=-\frac{17}{13}\Rightarrow \frac{-\pi }{2}<2\theta <0\Rightarrow 2(\frac{\pi }{2}+2\theta )+3(\frac{\pi }{2}-2\theta )+2\theta =\frac{\lambda \pi }{2}\Rightarrow \frac{5\pi }{2}=\frac{\lambda \pi }{2}\Rightarrow \lambda =5$