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

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

Explanation

Solution

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

Complete step by step answer:
Here, we are given a trigonometric function as sin⁡(cot⁡−1x)\sin \left( {{\cot }^{-1}}x \right). We have to find its value. For this, let us first suppose that cot⁡−1x=θ{{\cot }^{-1}}x=\theta . Now, we need to find the value of sin⁡θ\sin \theta where θ=cot⁡−1x\theta ={{\cot }^{-1}}x.
Multiplying cot on both sides of θ=cot⁡−1x\theta ={{\cot }^{-1}}x we get:
⇒cot⁡θ=cot⁡(cot⁡−1x)\Rightarrow \cot \theta =\cot \left( {{\cot }^{-1}}x \right)
As we know, aa−1=1a{{a}^{-1}}=1 therefore, cot and cot⁡−1{{\cot }^{-1}} cancels out at right side and we get:
⇒cot⁡θ=x\Rightarrow \cot \theta =x
Now let us use this to find the value of sin⁡θ\sin \theta in terms of x. As we know, cot⁡θ=cos⁡θsin⁡θ\cot \theta =\dfrac{\cos \theta }{\sin \theta } therefore above equation becomes,
⇒cos⁡θsin⁡θ=x\Rightarrow \dfrac{\cos \theta }{\sin \theta }=x
Since, cos⁡θ\cos \theta needs to be changed to sin⁡θ\sin \theta to get value sin⁡θ\sin \theta in terms of x, but it is possible only when we use squared terms. Hence, squaring both sides we get:
⇒cos⁡2θsin⁡2θ=x2\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }=x^2
Now we know that sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1. Taking sin⁡2θ{{\sin }^{2}}\theta on left side we get cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta . Let us put this value in above equation to get equation in terms of sin⁡θ\sin \theta and x only, we get:
⇒1−sin⁡2θsin⁡2θ=x\Rightarrow 1-\dfrac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }=x
Cross multiplying we get:
⇒1−sin⁡2θ=x2sin⁡2θ\Rightarrow 1-{{\sin }^{2}}\theta ={{x}^{2}}{{\sin }^{2}}\theta
Take sin⁡2θ{{\sin }^{2}}\theta terms one side and constant terms on other side, we get:
⇒1=x2sin⁡2θ+sin⁡2θ\Rightarrow 1={{x}^{2}}{{\sin }^{2}}\theta +{{\sin }^{2}}\theta
Taking sin⁡2θ{{\sin }^{2}}\theta common on right side, we get:
⇒1=(1+x2)sin⁡2θ\Rightarrow 1=\left( 1+{{x}^{2}} \right){{\sin }^{2}}\theta
Dividing both sides by (1+x2)\left( 1+{{x}^{2}} \right) and rearranging the sides, we get:
⇒sin⁡2θ=11+x2\Rightarrow {{\sin }^{2}}\theta =\dfrac{1}{1+{{x}^{2}}}
We need value of sin⁡θ\sin \theta so let us take square root on both sides, we get:

& \Rightarrow \sin \theta =\sqrt{\dfrac{1}{1+{{x}^{2}}}} \\\ & \Rightarrow \sin \theta =\pm \dfrac{1}{\sqrt{1+{{x}^{2}}}} \\\ \end{aligned}$$ 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.