Question

Write the set of values of x for which 2tan inverse x= tan inverse 2x/1-x^2

Answer 1

∴ $2tan^{-1}x=tan^{-1}\frac{2x}{1-x^{2} }$ when $-1< x< 1$

Step-by-step explanation:

Given;

$2tan^{-1}x=tan^{-1}\frac{2x}{1-x^{2} }$__equation-1

Let, $x=tanA$  Where $A=tan^{-1}x$

Now,

$\frac{2x}{1-x^{2} } =\frac{2tanA}{1-tan^{2} A}$

⇒$\frac{2x}{1-x^{2} } =tan2A$

take $tan^{-1}$ on both sides;

⇒$tan^{-1} \frac{2x}{1-x^{2} } =tan^{-1}tan2A$

We know,

$tan^{-1} (tan\theta) =\theta$ when $-\frac{\pi}{2}< \theta < \frac{\pi}{2}$

⇒$tan^{-1} (tan2A) =2A$

When,

$-\frac{\pi}{2}< 2A < \frac{\pi}{2}$

⇒$-\frac{\pi}{4}< A < \frac{\pi}{4}$

⇒$-\frac{\pi}{4}< tan^{-1} x < \frac{\pi}{4}$

⇒$tan^{-1} (-\frac{\pi}{4})<x < tan^{-1} (\frac{\pi}{4})$

⇒$-1< x< 1$

Hence $2tan^{-1}x=tan^{-1}\frac{2x}{1-x^{2} }$ when $-1< x< 1$