Question

If arcsin[2a/(1+a^2)] + arccos[(1 - a^2)/(1 + a^2)] = arctan(2x/(1-x^2)) where a, x ∈ (0,1), then the value of x is__?​

Answer 1

Inverse Trigonometric Functions

Formulae:

$\quad\mathsf{2tan^{-1}x=sin^{-1}\left(\frac{2x}{1-x^{2}}\right)}$

$\quad\quad\quad\mathsf{=tan^{-1}\left(\frac{2x}{1-x^{2}}\right)}$

$\quad\quad\quad\mathsf{=cos^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)}$

To find: Value of $\mathsf{x}$ is ___ .

Solution:

$\therefore \mathsf{sin^{-1}\left[\frac{2a}{1+a^{2}}\right]+cos^{-1}\left[\frac{1-a^{2}}{1+a^{2}}\right]=tan^{-1}\left[\frac{2x}{1-x^{2}}\right]}$

$\Rightarrow \mathsf{2tan^{-1}a+2tan^{-1}a=2tan^{-1}x}$

$\Rightarrow \mathsf{2tan^{-1}a=tan^{-1}x}$

$\Rightarrow \mathsf{tan^{-1}x=2tan^{-1}a}$

$\Rightarrow \mathsf{tan^{-1}x=tan^{-1}\left[\frac{2a}{1-a^{2}}\right]}$

$\Rightarrow \color{blue}{\mathsf{x=\frac{2a}{1-a^{2}}}}$

Answer: $\color{blue}{\bold{x=\frac{2a}{1-a^{2}}}}$

Note: The problems of 'Inverse Trigonometric Functions' can be solved easily if the formulae are known to you. However if is is difficult in conversions, consider $\mathsf{x=r\:cos\theta}$ or $\mathsf{r\:sin\theta}$, etc.