Inverse Circular Functions 2 Question 6

6. The value of $x$ for which $\sin \left[\cot ^{-1}(1+x)\right]=\cos \left(\tan ^{-1} x\right)$ is

(a) $\frac{1}{2}$

(b) 1

(c) 0

(d) $-\frac{1}{2}$

(2004, 1M)

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Answer:

Correct Answer: 6. (d)

Solution:

  1. Given, $\sin \left[\cot ^{-1}(1+x)\right]=\cos \left(\tan ^{-1} x\right)$

and we know that,

$\cot ^{-1} \theta=\sin ^{-1} \frac{1}{\sqrt{1+\theta^{2}}}$ and $\tan ^{-1} \theta=\cos ^{-1} \frac{1}{\sqrt{1+\theta^{2}}}$

From Eq. (i),

$ \begin{aligned} & & \sin \sin ^{-1} \frac{1}{\sqrt{1+(1+x)^{2}}} & =\cos \cos ^{-1} \frac{1}{\sqrt{1+x^{2}}} \\ \Rightarrow & & \frac{1}{\sqrt{1+(1+x)^{2}}} & =\frac{1}{\sqrt{1+x^{2}}} \\ \Rightarrow & & 1+x^{2}+2 x+1 & =x^{2}+1 \\ \Rightarrow & & x & =-\frac{1}{2} \end{aligned} $



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