To express [tex]\(\sin \left(\cot ^{-1}(x)-\cot ^{-1}(y)\right)\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] only, we will use the following trigonometric identities:
1. [tex]\(\cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right)\)[/tex].
First, let's rewrite the given expression using this identity:
[tex]\[
\sin \left(\cot ^{-1}(x)-\cot ^{-1}(y)\right) = \sin \left(\tan ^{-1}\left(\frac{1}{x}\right) - \tan ^{-1}\left(\frac{1}{y}\right)\right)
\][/tex]
Next, use the addition formula for sine:
[tex]\[
\sin(a - b) = \sin(a) \cos(b) - \cos(a) \sin(b)
\][/tex]
Set [tex]\(a = \tan^{-1}\left(\frac{1}{x}\right)\)[/tex] and [tex]\(b = \tan^{-1}\left(\frac{1}{y}\right)\)[/tex]. We need to find [tex]\(\sin(a)\)[/tex], [tex]\(\cos(a)\)[/tex], [tex]\(\sin(b)\)[/tex], and [tex]\(\cos(b)\)[/tex].
For [tex]\(a = \tan^{-1}\left(\frac{1}{x}\right)\)[/tex]:
[tex]\[
\tan(a) = \frac{1}{x} \implies a = \tan^{-1}\left(\frac{1}{x}\right)
\][/tex]
[tex]\[
\sin(a) = \frac{\frac{1}{x}}{\sqrt{1+\left(\frac{1}{x}\right)^2}} = \frac{1/x}{\sqrt{1+1/x^2}} = \frac{1}{\sqrt{x^2 + 1}}
\][/tex]
[tex]\[
\cos(a) = \frac{1}{\sqrt{1+\left(\frac{1}{x}\right)^2}} = \frac{1}{\sqrt{x^2 + 1}}
\][/tex]
For [tex]\(b = \tan^{-1}\left(\frac{1}{y}\right)\)[/tex]:
[tex]\[
\tan(b) = \frac{1}{y} \implies b = \tan^{-1}\left(\frac{1}{y}\right)
\][/tex]
[tex]\[
\sin(b) = \frac{\frac{1}{y}}{\sqrt{1+\left(\frac{1}{y}\right)^2}} = \frac{1/y}{\sqrt{1+1/y^2}} = \frac{1}{\sqrt{y^2 + 1}}
\][/tex]
[tex]\[
\cos(b) = \frac{1}{\sqrt{1+\left(\frac{1}{y}\right)^2}} = \frac{1}{\sqrt{y^2 + 1}}
\][/tex]
Substitute these values into the sine difference formula:
[tex]\[
\sin \left(\tan ^{-1}\left(\frac{1}{x}\right) - \tan ^{-1}\left(\frac{1}{y}\right)\right) = \sin(a) \cos(b) - \cos(a) \sin(b)
\][/tex]
[tex]\[
= \frac{1}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{y^2 + 1}} - \frac{1}{\sqrt{x^2 + 1}} \cdot \frac{1}{\sqrt{y^2 + 1}}
\][/tex]
Since both terms are identical, they cancel each other out, simplifying to:
[tex]\[
= 0
\][/tex]
Thus, the expression [tex]\(\sin \left(\cot ^{-1}(x)-\cot ^{-1}(y)\right)\)[/tex] in terms of [tex]\(x\)[/tex] and [tex]\(y\)[/tex] simplifies to:
[tex]\[
\boxed{\sin \left(\tan^{-1}\left(\frac{1}{x}\right) - \tan^{-1}\left(\frac{1}{y}\right)\right)}
\][/tex]
