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Question

Question: \[\tan\lbrack\sec^{- 1}\sqrt{1 + x^{2}}\rbrack =\]...

tan⁡[sec⁡−11+x2]=\tan\lbrack\sec^{- 1}\sqrt{1 + x^{2}}\rbrack =

A

1x\frac{1}{x}

B

x

C

11+x2\frac{1}{\sqrt{1 + x^{2}}}

D

x1+x2\frac{x}{\sqrt{1 + x^{2}}}

Answer

x

Explanation

Solution

y = tan[sec⁡−11+x2]\left\lbrack \sec^{- 1}\sqrt{1 + x^{2}} \right\rbrack let sec–11+x2\sqrt{1 + x^{2}} = q

̃ sec q = 1+x2\sqrt{1 + x^{2}},  Now y = tan q, y = x