Question

Write the trigonometric expression as an algebraic expression containing u and v

$\sin \left({\tan }^{-1}u-{\sin }^{-1}v\right)$

Short answer

$\sin \left({\tan }^{-1}u-{\sin }^{-1}v\right)=\frac{u\sqrt{1-v^2}-v}{\sqrt{u^2+1}}$

Step-by-step solution

Step 1. Given information we have is

a trigonometric equation$\sin \left({\tan }^{-1}u-{\sin }^{-1}v\right)$

Step 2. putting tan-1u=xu=tanxsin-1v=yv=siny

$u=\frac{\sin u}{\cos u}$

$u\cos u=\sin u$

Step 3. Using the Pythagoras theoremsin2x+cos2x=1

${\left(u\cos x\right)}^2+{\cos }^{2}x=1$

$$\begin{gathered} u^2{\cos }^{2}x+{\cos }^{2}x=1 \\ {\cos }^{2}x(u^2+1)=1 \\ {\cos }^{2}x=\frac{1}{u^2+1} \\ \cos x=\frac{1}{\sqrt{u^2+1}} \end{gathered}$$

Now,

$\sin x=\frac{u}{\sqrt{u^2+1}}$

Step 4. Now, we know thatcosy=1-sin2ycosy=1-v2using the formula ofsinx-y=sinxcosy-cosxsiny we get

On Solving

$$\begin{gathered} =\frac{u}{\sqrt{u^2+1}}\sqrt{1-v^2}-\frac{v}{\sqrt{u^2+1}} \\ =\frac{u\sqrt{1-v^2}-v}{\sqrt{u^2+1}} \end{gathered}$$