Chapter 7.2, Problem 45E

To determine

To write: the expression in terms of x and y only.

Answer to Problem 45E

  $\sin \left({\tan }^{-1}x-{\tan }^{-1}y\right)=\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}$ .

Explanation of Solution

Given information:

Given expression,

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

Formula used:

Subtraction formula of sine

  $\sin \left(u-v\right)=\sin u\cos v-\cos u\sin v$

Calculation:

Now, consider the given expression

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

Now, let us take $u={\tan }^{-1}x,v={\tan }^{-1}y$ .

So, using subtraction formula, we get

  $\begin{array}{c}\sin \left(u-v\right)=\sin u\cos v-\cos u\sin v\\ \sin \left({\tan }^{-1}x-{\tan }^{-1}y\right)=\sin \left({\tan }^{-1}x\right)\cos \left({\tan }^{-1}y\right)-\cos \left({\tan }^{-1}x\right)\sin \left({\tan }^{-1}y\right)\end{array}$ …… (1)

Now, $u={\sin }^{-1}x$

  $\begin{array}{c}\tan u=x\\ \sin u=\frac{1}{\sqrt{1+{\cot }^{2}u}}\\ =\frac{1}{\sqrt{1+\frac{1}{x^2}}}\\ \sin \left({\tan }^{-1}x\right)=\frac{x}{\sqrt{1+x^2}}\\ \cos \left({\tan }^{-1}x\right)=\sqrt{1-{\left(\frac{x}{\sqrt{1+x^2}}\right)}^2}\\ =\sqrt{\frac{1}{1+x^2}}\\ =\frac{1}{\sqrt{1+x^2}}\end{array}$

Further $v={\tan }^{-1}y$

  $\begin{array}{c}\tan v=y\\ \cos v=\frac{1}{\sqrt{1+{\tan }^{2}v}}\\ =\frac{1}{\sqrt{1+y^2}}\\ \cos \left({\tan }^{-1}y\right)=\frac{1}{\sqrt{1+y^2}}\\ \sin \left({\tan }^{-1}y\right)=\sqrt{1-{\left(\frac{1}{\sqrt{1+y^2}}\right)}^2}\\ =\sqrt{1-\frac{1}{1+y^2}}\\ =\frac{y}{\sqrt{1+y^2}}\end{array}$

Now, substitute the values in equation (1).

  $\begin{array}{c}\sin \left({\tan }^{-1}x-{\tan }^{-1}y\right)=\sin \left({\tan }^{-1}x\right)\cos \left({\tan }^{-1}y\right)-\cos \left({\tan }^{-1}x\right)\sin \left({\tan }^{-1}y\right)\\ =\frac{x}{\sqrt{1+x^2}}\times \frac{1}{\sqrt{1+y^2}}-\frac{1}{\sqrt{1+x^2}}\times \frac{y}{\sqrt{1+y^2}}\\ =\frac{x-y}{\sqrt{1+x^2}\sqrt{1+y^2}}\end{array}$