Question

If $\tan^{-1} \frac{\sqrt{1+x^2}-1}{x} = 4^\circ$, then;

• A. x = tan 2°
• B. x = tan 4°
• C. x = tan $(1/4)^\circ$
• D. x = tan 8°

Answer 1

Answer: (D)

x = tan 8°

Answer 2

Given:

$\tan^{-1}\left( \frac{\sqrt{1+x^2}-1}{x} \right)= 4^\circ$.

Let $\theta = 4^\circ$. Then,

$\frac{\sqrt{1+x^2}-1}{x} = \tan 4^\circ$.

Recall the half-angle identity:

$\tan\frac{\phi}{2} = \frac{\sqrt{1+\tan^2\phi}-1}{\tan\phi}$.

Setting $x = \tan\phi$, we have:

$\frac{\sqrt{1+\tan^2\phi}-1}{\tan\phi} = \tan\frac{\phi}{2}$.

Comparing with the given expression, we require:

$\tan\frac{\phi}{2} = \tan4^\circ \quad \Longrightarrow \quad \frac{\phi}{2} = 4^\circ \quad \Longrightarrow \quad \phi = 8^\circ$.

Thus, $x = \tan\phi = \tan 8^\circ$.