Question

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Answer 1

Answer:

$\displaystyle x = \frac{1}{3}$.

Step-by-step explanation:

Consider the double angle identity for tangents:

$\displaystyle \tan(2\theta) = \frac{2\tan{\theta}}{1 - \tan^{2}{\theta}}$.

Also, for two complementary angles,

$\displaystyle \tan{\left(\frac{\pi}{2} - \theta \right)} = \frac{1}{\tan{\theta}}$.

Subtract $\tan^{-1}(x + 1)$ from both sides of this equation:

$\displaystyle 2\tan^{-1}{x} = \frac{\pi}{2} + (-\tan^{-1}(x + 1))$.

Take the tangent of both sides of this equation:

$\displaystyle \tan(2\tan^{-1}{x}) = \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right)$.

Apply the double-angle identity to the left-hand side of this equation:

$\displaystyle \tan(2\tan^{-1}{x}) \implies \frac{2\tan(\tan^{-1}x)}{1 - \tan^{2}(\tan^{-1}x)}\implies \frac{2x}{1 - x^{2}}$.

The two angles $\displaystyle \left(\frac{\pi}{2} + (-\tan^{-1}\left(x + 1\right))\right)$ and $(x - 1)$ are complementary. Therefore, for the right-hand side of this equation,

$\displaystyle \tan\left(\frac{\pi}{2} -\tan^{-1}\left(x + 1\right)\right)\implies \frac{1}{\tan(\tan^{-1}(x + 1))} \implies \frac{1}{x + 1}$.

Equate the two sides of this equation:

$\begin{aligned} & \frac{2x}{1 - x^{2}} = \frac{1}{x + 1}\\ \implies & \frac{2x}{(1 - x)(1 + x)} = \frac{1}{1 + x}\\\implies & 2x = 1 - x,\; x \ne 1,\; x \ne -1\\\implies & x = \frac{1}{3}\end{aligned}$.