Solveeit Logo
Login

Question

Mathematics Question on Inverse Trigonometric Functions

If 0<x<10 < x < 1, then \sqrt{1+x^2}[\\{x cos(cot^{-1}x) +sin(cot^{-1}x)\\}^2-1]^{1/2} is equal to

A

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

B

xx

C

x 1+x2x\,\sqrt{1+x^2}

D

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

Answer

x 1+x2x\,\sqrt{1+x^2}

Explanation

Solution

We have, 0 < x < 1
Let          cot−1x=θ \, \, \, \, \, \, \, \, \, cot^{-1}x=\theta
⇒            cotθ=x\Rightarrow \, \, \, \, \, \, \, \, \, \, \, \, cot \theta=x
⇒           sinθ=11+x2=sin(cot−1x)\Rightarrow \, \, \, \, \, \, \, \, \, \, \, sin\theta=\frac{1}{\sqrt{1+x^2}}=sin(cot^{-1}x)
and             cosθ=x1+x2=cos(cot−1x) \, \, \, \, \, \, \, \, \, \, \, \, cos \theta =\frac{x}{\sqrt{1+x^2}}=cos(cot^{-1}x)
Now, 1+x2[xcos(cot−1x)+sin(cot−1x)2−1]1/2Now, \, \sqrt{1+x^2}[\\{x cos (cot^{-1}x)+sin(cot^{-1}x)\\}^2 -1]^{1/2}
           =1+x2[(xx1+x2+11+x2)2−1]1/2\, \, \, \, \, \, \, \, \, \, \, =\sqrt{1+x^2}\bigg[\bigg(x\frac{x}{\sqrt1+x^2}+\frac{1}{\sqrt{1+x^2}}\bigg)^2-1\bigg]^{1/2}
           =1+x2[(x1+x21+x2)2−1]1/21\, \, \, \, \, \, \, \, \, \, \, =\sqrt{1+x^2}\bigg[\bigg(x\frac{1+x^2}{\sqrt1+x^2}\bigg)^2-1\bigg]^{1/21}
           =1+x2[1+x2−1]1/2=x1+x2\, \, \, \, \, \, \, \, \, \, \, =\sqrt{1+x^2}[1+x^2-1]^{1/2}=x\sqrt{1+x^2}