Inverse Circular Functions 2 Question 5
5. If $0<x<1$, then $\sqrt{1+x^{2}}\left[{x \cos \left(\cot ^{-1} x\right)\right.\right.$ $\left.\left.+\sin \left(\cot ^{-1} x\right) }^{2}-1\right]^{1 / 2}$ is equal to
(2008, 3M)
(a) $\frac{x}{\sqrt{1+x^{2}}}$
(b) $x$
(c) $x \sqrt{1+x^{2}}$
(d) $\sqrt{1+x^{2}}$
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Answer:
Correct Answer: 5. (c)
Solution:
- We have, $0<x<1$
Let $\quad \cot ^{-1} x=\theta$
$ \begin{array}{ll} \Rightarrow & \cot \theta=x \\ \Rightarrow & \sin \theta=\frac{1}{\sqrt{1+x^{2}}}=\sin \left(\cot ^{-1} x\right) \\ \text { and } & \cos \theta=\frac{x}{\sqrt{1+x^{2}}}=\cos \left(\cot ^{-1} x\right) \end{array} $
Now, $\sqrt{1+x^{2}}\left[{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right) }^{2}-1\right]^{1 / 2}$
$ \begin{aligned} & =\sqrt{1+x^{2}} \quad x \frac{x}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+x^{2}}}-1 \\ & =\sqrt{1+x^{2}} \quad \frac{1+x^{2}}{\sqrt{1+x^{2}}}-1 \\ & =\sqrt{1+x^{2}}\left[1+x^{2}-1\right]^{1 / 2}=x \sqrt{1+x^{2}} \end{aligned} $
