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:
- 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} $