Question

Write the trigonometric expression as an algebraic expression containing uand v

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

Short answer

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

Step-by-step solution

Step 1. Given information 

a trigonometric equation$sec({\tan }^{-1}u+{\cos }^{-1}v)$

Step 2. Putting tan-1u=xu=tanxcos-1v=yv=cosy

We know that,

$secx=\sqrt{1+{\tan }^{2}x}$

also, $secx=\sqrt{1+u^2}$

using the reciprocal identity of secx we get

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

Step 3. Using identity,

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

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

Also,

$\sin y=\sqrt{1-v^2}$

Step 4. Using the formula cosx+y=cosxcosy-sinxsiny

$$\begin{gathered} \frac{1}{\cos \left(x+y\right)}=\frac{1}{v\sqrt{1+u^2}-\left(u\sqrt{1+u^2}\right)\left(\sqrt{1-v^2}\right)} \\ \frac{1}{\cos \left(x+y\right)}=\frac{1}{\sqrt{1+u^2}(v-u\sqrt{1-v^2})} \end{gathered}$$