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← IIT-JEE 2008 · Paper 1
MathematicsSingle correct+3 / -1Question 2 of 69

IIT-JEE 2008 · Paper 1, Mathematics question 2

If 0<x<10<x<1, then
1+x2[{xcos⁡(cot⁡−1x)+sin⁡(cot⁡−1x)}2−1]1/2=\sqrt{1+x^2}\left[\{x\cos(\cot^{-1}x)+\sin(\cot^{-1}x)\}^2-1\right]^{1/2}=

Choose one option
Answer and worked solution

Official answer: C

Let θ=cot⁡−1x\theta=\cot^{-1}x. Then cos⁡θ=x1+x2\cos\theta=\dfrac x{\sqrt{1+x^2}}, sin⁡θ=11+x2\sin\theta=\dfrac1{\sqrt{1+x^2}}, so
xcos⁡θ+sin⁡θ=x2+11+x2=1+x2.x\cos\theta+\sin\theta=\frac{x^2+1}{\sqrt{1+x^2}}=\sqrt{1+x^2}.

Simplify. [(1+x2)−1]1/2=x\left[(1+x^2)-1\right]^{1/2}=x (as x>0x>0), and the expression is x1+x2x\sqrt{1+x^2}.

(C)

Topics:Inverse trigonometric functions
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