Question

The value of $\sin \left( {{\cot }^{-1}}x \right)$ is:
a) $\sqrt{1+{{x}^{2}}}$
b) $x$
c) ${{\left( 1+{{x}^{2}} \right)}^{-\dfrac{3}{2}}}$
d) ${{\left( 1+{{x}^{2}} \right)}^{-\dfrac{1}{2}}}$

Answer 1

Hint: To start with the solution, let ${{\cot }^{-1}}x$ to be equal to y and express y in terms of inverse of sine. Finally, use the property that $\sin \left( {{\sin }^{-1}}p \right)=p$ , provided p is a positive number less than or equal to 1.

Complete step-by-step answer:

To start with the question, we let ${{\cot }^{-1}}x$ to be equal to y.

$\therefore y={{\cot }^{-1}}x$

$\Rightarrow \cot y=x$

Now we know that $\cot y=\dfrac{base}{perpendicular}=x$. Therefore, we can say that the base is equal to xk and perpendicular is equal to k, where k is a natural number.

We can also deduce that the length of the hypotenuse is equal to $\sqrt{{{\left( base \right)}^{2}}+{{\left( perpendicular \right)}^{2}}}=\sqrt{{{x}^{2}}{{k}^{2}}+{{k}^{2}}}$

$\therefore \sin y=\dfrac{perpendicular}{hypotenuse}$

$\Rightarrow \sin y=\dfrac{k}{\sqrt{{{k}^{2}}{{x}^{2}}+{{k}^{2}}}}$

$\Rightarrow \sin y=\dfrac{k}{k\sqrt{{{x}^{2}}+1}}$

$\Rightarrow \sin y=\dfrac{1}{\sqrt{{{x}^{2}}+1}}$

$\Rightarrow y={{\sin }^{-1}}{{\left( 1+{{x}^{2}} \right)}^{-\dfrac{1}{2}}}$

Now we will substitute y in the expression given in the question. On doing so, we get

$\sin \left( {{\cot }^{-1}}x \right)$

$=\sin \left( {{\sin }^{-1}}{{\left( \sqrt{1+{{x}^{2}}} \right)}^{-\dfrac{1}{2}}} \right)$

Here we can clearly see that ${{\left( \sqrt{1+{{x}^{2}}} \right)}^{-\dfrac{1}{2}}}$ is positive and will always be less than or equal to one as the denominator will be greater than equal to 1, and the numerator is equal to 1. . So, we can use the formula $\sin \left( {{\sin }^{-1}}p \right)=p$ . On doing so, our expression becomes:

${{\left( \sqrt{1+{{x}^{2}}} \right)}^{-\dfrac{1}{2}}}$

Therefore, the answer to the above question is option (d).

Note: While dealing with inverse trigonometric functions, it is preferred to know about the domains and ranges of the different inverse trigonometric functions. For example: the domain of ${{\sin }^{-1}}x$ is $[-1,1]$ and the range is $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.