Question

Write the trigonometric expression as an algebraic expression containing uand v

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

Short answer

$=\frac{uv-\sqrt{1-u^2}\sqrt{1-v^2}}{v\sqrt{1-u^2}+u\sqrt{1-v^2}}$

Step-by-step solution

Step 1. The given information 

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

Step 2. Putting sin-1u=xu=sinxcos-1v=yv=cosy

We know that,

$\cos x=\sqrt{1-{\sin }^{2}x}$

now,

$\cos x=\sqrt{1-u^2}$

Also, $\tan x=\frac{\sin x}{\cos x}$

$=\frac{u}{\sqrt{1-u^2}}$

Step 3. Similarly

$$\begin{gathered} \sin y=\sqrt{1-v^2} \\ \tan y=\frac{\sqrt{1-v^2}}{v} \end{gathered}$$

Step 4. Using the formula tanx-y=tanx-tany1+tanxtany

Solving we get,

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

$\frac{uv-\sqrt{(1-u^2)(1-v^2)}}{v\sqrt{1-u^2}+u\sqrt{1-v^2}}$