Question

Write the trigonometric expression as an algebraic expression containing uand v

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

Short answer

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

Step-by-step solution

Step 1. Given information is

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

putting

$$\begin{gathered} {\tan }^{-1}u=x \\ u=\tan x \\ {\tan }^{-1}v=y \\ v=\tan y \end{gathered}$$

Step 2. Using trigonometry identity

$$\begin{gathered} \frac{1}{\cos x}=\sqrt{1+{\tan }^{2}x} \\ \frac{1}{\cos x}=\sqrt{1+u^2} \\ \tan x=\frac{\sin x}{\cos x} \\ \sin x=u\sqrt{1+v^2} \end{gathered}$$

Step 3.  Using the identity we get,

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

Step 4. Using the formulacosx+y=cosxcosy-sinxsiny

$$\begin{gathered} =\left(\sqrt{1+u^2}\right)\left(\sqrt{1+v^2}\right)-\left(u\sqrt{1+u^2}\right)\left(v\sqrt{1+v^2}\right) \\ =\left(\sqrt{1+u^2}\right)\left(\sqrt{1+v^2}\right)(1-uv) \end{gathered}$$On solving we get