Question

Find the value of $x$ for which $\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)$ is satisfied.
(a) $\dfrac{1}{2}$
(b) 1
(c) 0
(d) $-\dfrac{1}{2}$

Answer 1

In this question, we are given with the equation $\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)$.
Now we will use the identity of the trigonometric function say $\sin \left( \theta \right)=\cos \left( \dfrac{\pi }{2}-\theta \right)$. Also we will use the property that if $\cos x=\cos y$, then we have that $x=2n\pi \pm y$ for integers values $n$. Using these properties of trigonometric functions, we will get our desired answer.

Complete step by step answer:
We are given with the equation $\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right).............(1)$.
Now since we know that for any $\theta \in \left[ 0,2\pi \right]$, $\sin \left( \theta \right)=\cos \left( \dfrac{\pi }{2}-\theta \right)$.
Therefore we can have
$\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( \dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right) \right).............(2)$
Now on substituting the value of equation (2) in equation (1), we will have
$\cos \left( \dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right).............(3)$
Since we also know the property of the trigonometric function that if $\cos a=\cos b$, then we have that $a=2n\pi \pm b$ for all integers values $n$.
Now on comparing equation (3) with $\cos a=\cos b$, we have that
$a=\dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right)$ and
$b={{\tan }^{-1}}x$
Therefore using the property that if $\cos a=\cos b$, then we have that $a=2n\pi \pm b$ for all integers values $n$, we will have
$\dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right)=2n\pi \pm {{\tan }^{-1}}x$ for all integers values $n$.
Now since the above equation is true all the integer values of $n$, hence it is also true for $n=0$.
Now on substituting the value $n=0$ in the above equation $\dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right)=2n\pi \pm {{\tan }^{-1}}x$, we will have
$\dfrac{\pi }{2}-{{\cot }^{-1}}\left( 1+x \right)=0\pm {{\tan }^{-1}}x$
This implies that
$\dfrac{\pi }{2}={{\cot }^{-1}}\left( 1+x \right)\pm {{\tan }^{-1}}x............(4)$
Now since we know that ${{\cot }^{-1}}x=\dfrac{\pi }{2}-{{\tan }^{-1}}x$ for all values of $x$.
Hence we can have ${{\cot }^{-1}}\left( 1+x \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( 1+x \right)$
Therefore on substituting the value ${{\cot }^{-1}}\left( 1+x \right)=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( 1+x \right)$ in the above equation (4), we get
$\dfrac{\pi }{2}=\dfrac{\pi }{2}-{{\tan }^{-1}}\left( 1+x \right)\pm {{\tan }^{-1}}x$
Therefore we have
${{\tan }^{-1}}\left( 1+x \right)=\pm {{\tan }^{-1}}x$
Now since we know that if $\pm {{\tan }^{-1}}x={{\tan }^{-1}}\left( \pm x \right)$
Therefore we get
${{\tan }^{-1}}\left( 1+x \right)={{\tan }^{-1}}\left( \pm x \right)$
Also if ${{\tan }^{-1}}x={{\tan }^{-1}}y$, then $x=y$.
Therefore form ${{\tan }^{-1}}\left( 1+x \right)={{\tan }^{-1}}\left( \pm x \right)$, we get that
$1+x=\pm x$
Now since we cannot have $1+x=x$ for any value of $x$, hence we must have
$1+x=-x$
This implies that

& -2x=1 \\\ & \Rightarrow x=-\dfrac{1}{2} \end{aligned}$$ Therefore we have that for $$x=-\dfrac{1}{2}$$ we have that the given equation $$\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)$$ is satisfied. **So, the correct answer is “Option D”.** **Note:** In this problem, in order to determine the value of $$x$$ for which the given equation $$\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)$$ is satisfied we can substitute the value $${{\cot }^{-1}}\left( 1+x \right)=a$$ and $${{\tan }^{-1}}x=b$$ to get that $$1+x=\cot a$$ and $$x=\tan b$$. Then we have found the values of $$a$$ and $$b$$ using the properties of $$\cot $$ and $$\tan $$ function. Then on substituting the values in $$\sin \left( {{\cot }^{-1}}\left( 1+x \right) \right)=\cos \left( {{\tan }^{-1}}x \right)$$ we have form a quadratic equation in $$x$$. On solving the same we can get the desired answer.