If tan−1⁡1+x2−1x=40, then

1+x2−1x=tan⁡40⇒1+x2=2xtan⁡40+1+x2tan2⁡40⇒x=0 or x=2tan⁡401+tan2⁡40=tan80Since x≠0, we have x=tan⁡80

If tan−1⁡1+x2−1x=40, then

1+x2−1x=tan⁡40⇒1+x2=2xtan⁡40+1+x2tan2⁡40⇒x=0 or x=2tan⁡401+tan2⁡40=tan80Since x≠0, we have x=tan⁡80

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If $\tan^{- 1} \frac{\sqrt{1 + x^{2}} - 1}{x} = 4^{0}$ , then

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x = tan2°

x = tan4°

$x = \tan (1 / 4)^{0}$

$x = \tan 8^{0}$

answer is D.

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Detailed Solution

$\begin{array}{l} \frac{\sqrt{1 + x^{2}} - 1}{x} = \tan 4^{0} \Rightarrow 1 + x^{2} = 2 xtan 4^{0} + 1 + x^{2} \tan^{2} 4^{0} \\ \Rightarrow x = 0 or x = \frac{2 \tan 4^{0}}{1 + \tan^{2} 4^{0}} = \tan 8^{0} \end{array}$
Since $x \neq 0, we have x = \tan 8^{0}$