Question

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

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $0$
• D. $\frac{9}{4}$

Answer 1

The correct option is A

$-\frac{1}{2}$

Explanation for the correct option.

Step 1: Evaluate the expression

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

Converting $co{t}^{-1}$in terms of ${\sin }^{-1}$ and ${\tan }^{-1}$ in terms of ${\cos }^{-1}$ we get:

$\begin{array}{rcl}\sin \left[{\cot }^{-1}(x+1)\right]& =& \sin \left[{\sin }^{-1}\frac{1}{\sqrt{x^2+2x+2}}\right]\\ & =& \frac{1}{\sqrt{x^2+2x+2}}....\left(1\right)\\ \cos \left({\tan }^{-1}x\right)& =& \cos \left[{\cos }^{-1}\frac{1}{\sqrt{1+x^2}}\right]\\ & =& \frac{1}{\sqrt{1+x^2}}...\left(2\right)\end{array}$

Step 2: Solve for $x$

Equating (1) & (2) we get:

$\begin{array}{rcl}\frac{1}{\sqrt{x^2+2x+2}}& =& \frac{1}{\sqrt{1+x^2}}\\ & \Rightarrow & \sqrt{1+x^2}=\sqrt{x^2+2x+2}\\ & \Rightarrow & 1+x^2=x^2+2x+2\\ & \Rightarrow & 2x+1=0\\ & \Rightarrow & x=-\frac{1}{2}\end{array}$

Hence, option A is correct.