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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 7: Q. 90 (page 487)

Write the trigonometric expression as an algebraic expression containing uand v
$s e c (\tan^{- 1} u + \cos^{- 1} v)$

Short Answer

Expert verified

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

Step by step solution

01

Step 1. Given information

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

02

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

We know that,

$s e c x = \sqrt{1 + \tan^{2} x}$

also, $s e c x = \sqrt{1 + u^{2}}$

using the reciprocal identity of secx we get

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

03

Step 3. Using identity,

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

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

Also,

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

04

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

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

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Most popular questions from this chapter

$\sin^{2} \frac{胃}{2} = \frac{______}{2}$ .

Show that $\cos \frac{蟺}{8} = \frac{\sqrt{2 + \sqrt{2}}}{2}$ and use it to find $\sin \frac{蟺}{16}, \cos \frac{蟺}{16}$ .

Establish each identity.

$\frac{\sin 胃 + \cos 胃}{\sin 胃} - \frac{\cos 胃 - \sin 胃}{\cos 胃} = s e c 胃 c s c 胃$

$\tan \frac{胃}{2} = \frac{1 - \cos 胃}{_____}$ .

Establish each identity.

$\frac{1 + \sin 胃 + \cos 胃}{1 + \sin 胃 - \cos 胃} = \frac{1 + \cos 胃}{\sin 胃}$

See all solutions

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