$ = \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right) + \frac{1}{2}{{\cos }^{ - 1}}\left( {\frac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}} \right)} \right]$
(Let $a = \tan \theta $)
$ = \tan \left[ {\frac{1}{2}{{\sin }^{ - 1}}(\sin 2\theta ) + \frac{1}{2}{{\cos }^{ - 1}}(\cos 2\theta )} \right]$
$ = \tan (2\theta ) = \tan 2\theta = \frac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }} = \frac{{2a}}{{1 - {a^2}}}$
Trick : Put $a = 0$, then tan $(0+0)=0;$ which is given by $(a)$ and $(c)$.
Again put $a = 1$, then $\tan \left( {\frac{\pi }{4} + \frac{\pi }{4}} \right) = \infty $, which is given by $(c).$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If $8 x^{2}+b x+c=0$ is a quadratic equation whose roots are $\alpha^{1 / 5}$ and $\beta^{1 / 5}$, then the value of $c-b$ is equal to: