Question

The value of $\sin \left( {{\cot }^{-1}}x \right)$ is:

Answer 1

Hint: In this question, we have to find the value of $\sin \left( {{\cot }^{-1}}x \right)$. For this, we will first suppose $\left( {{\cot }^{-1}}x \right)$ to be equal to $\theta $ and then perform various trigonometric operations to find value of $\sin \theta $ which will give us value of $\sin \left( {{\cot }^{-1}}x \right)$. Since $\left( {{\cot }^{-1}}x \right)$ was supposed to be $\theta $. Trigonometric identities and operations that we will use are given as
(I) ${{\cot }^{-1}}x=\theta $ implies that $\cot \theta =x$.
(II) $\cot \theta $ can be written in the form of $\dfrac{\cos \theta }{\sin \theta }$.
(III) ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.

Complete step by step answer:
Here, we are given a trigonometric function as $\sin \left( {{\cot }^{-1}}x \right)$. We have to find its value. For this, let us first suppose that ${{\cot }^{-1}}x=\theta $. Now, we need to find the value of $\sin \theta $ where $\theta ={{\cot }^{-1}}x$.
Multiplying cot on both sides of $\theta ={{\cot }^{-1}}x$ we get:
\[\Rightarrow \cot \theta =\cot \left( {{\cot }^{-1}}x \right)\]
As we know, $a{{a}^{-1}}=1$ therefore, cot and ${{\cot }^{-1}}$ cancels out at right side and we get:
\[\Rightarrow \cot \theta =x\]
Now let us use this to find the value of $\sin \theta $ in terms of x. As we know, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ therefore above equation becomes,
\[\Rightarrow \dfrac{\cos \theta }{\sin \theta }=x\]
Since, $\cos \theta $ needs to be changed to $\sin \theta $ to get value $\sin \theta $ in terms of x, but it is possible only when we use squared terms. Hence, squaring both sides we get:
\[\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }=x^2\]
Now we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Taking ${{\sin }^{2}}\theta $ on left side we get ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $. Let us put this value in above equation to get equation in terms of $\sin \theta $ and x only, we get:
\[\Rightarrow 1-\dfrac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }=x\]
Cross multiplying we get:
\[\Rightarrow 1-{{\sin }^{2}}\theta ={{x}^{2}}{{\sin }^{2}}\theta \]
Take ${{\sin }^{2}}\theta $ terms one side and constant terms on other side, we get:
\[\Rightarrow 1={{x}^{2}}{{\sin }^{2}}\theta +{{\sin }^{2}}\theta \]
Taking ${{\sin }^{2}}\theta $ common on right side, we get:
\[\Rightarrow 1=\left( 1+{{x}^{2}} \right){{\sin }^{2}}\theta \]
Dividing both sides by $\left( 1+{{x}^{2}} \right)$ and rearranging the sides, we get:
\[\Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{1+{{x}^{2}}}\]
We need value of $\sin \theta $ so let us take square root on both sides, we get:
\[\begin{align}
  & \Rightarrow \sin \theta =\sqrt{\dfrac{1}{1+{{x}^{2}}}} \\
 & \Rightarrow \sin \theta =\pm \dfrac{1}{\sqrt{1+{{x}^{2}}}} \\
\end{align}\]
Hence, $\sin \theta $ can be equal to $\dfrac{1}{\sqrt{1+{{x}^{2}}}},\dfrac{-1}{\sqrt{1+{{x}^{2}}}}$.
Since $\left( {{\cot }^{-1}}x \right)$ was supposed to be $\theta $ then $\sin \theta $ becomes $\sin \left( {{\cot }^{-1}}x \right)\text{ and }\sin \left( {{\cot }^{-1}}x \right)=\dfrac{1}{\sqrt{1+{{x}^{2}}}},\sin \left( {{\cot }^{-1}}x \right)=\dfrac{-1}{\sqrt{1+{{x}^{2}}}}$.

Hence, we get our required result as:
Value of $\sin \left( {{\cot }^{-1}}x \right)$ is $\dfrac{1}{\sqrt{1+{{x}^{2}}}},\dfrac{-1}{\sqrt{1+{{x}^{2}}}}$.

Note: Students should learn all basic trigonometric identities and functions for solving these questions. While finding the square root of any squared term, do not forget the negative value. Both positive and negative values are required answers. Do all the steps carefully taking care of signs.