Multiple Angles in Terms of tan⁻¹x

Multiple Angles in Terms of tan⁻¹x

Multiple Angles in Terms of $${{\tan }^{-1}}x$$:

Let $${{\tan }^{-1}}x=A$$,

Where $$A\in \left( -\pi /2,\pi /2 \right)$$,

$$\tan 2A=\frac{2\operatorname{Tan}A}{1-{{\tan }^{2}}A}$$,

$$=\frac{2x}{1-{{x}^{2}}}$$ ,

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

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

2{{\tan }^{-1}}x=\left\{ \begin{align} & {{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right),\ \ if\ \ -1<x<1 \\ & \pi +{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right),\ \ if\ x>1 \\ & -\pi +{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right),\ \ if\ x<1 \\ \end{align} \right.,

Example: Find the $$\tan \left[ 2{{\tan }^{-1}}\left( \frac{\sqrt{1+{{x}^{2}}}-1}{x} \right) \right]$$

Solution:

Put x = tan θ

⇒ θ = tan⁻¹x

$$\tan \left[ 2{{\tan }^{-1}}\left( \frac{\sqrt{1+{{\tan }^{2}}\theta }-1}{\tan \theta } \right) \right]$$,

$$\tan \left[ 2{{\tan }^{-1}}\left( \frac{\sqrt{{{\sec }^{2}}\theta }-1}{\tan \theta } \right) \right]$$,

$$\tan \left[ 2{{\tan }^{-1}}\left( \frac{\sec \theta -1}{\tan \theta } \right) \right]$$,

$$\tan \left[ 2{{\tan }^{-1}}\left( \frac{1-\cos \theta }{\sin \theta } \right) \right]$$,

$$\tan \left[ 2{{\tan }^{-1}}\left( \frac{2{{\sin }^{2}}\left( \theta /2 \right)}{2\sin \theta /2\cos \theta /2} \right) \right]$$,

$$\tan \left[ 2{{\tan }^{-1}}\left( \frac{\sin \left( \theta /2 \right)}{\cos \theta /2} \right) \right]$$,

$$\tan \left[ 2{{\tan }^{-1}}\left( \tan \left( \theta /2 \right) \right) \right]$$,

$$\tan \left[ 2\times \theta /2 \right]=\tan \theta =x$$.