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Chapter 5: Problem 66

Verify the identity. $$\sec ^{-1} \frac{1}{x}+\csc ^{-1} \frac{1}{x}=\frac{\pi}{2}$$

Short Answer

Expert verified
The identity \(\sec ^{-1} \frac{1}{x}+\csc ^{-1} \frac{1}{x}=\frac{\pi}{2}\) has been proven to be true.

Step by step solution

01

- Write in terms of cosine and sine

The first step is to write the secant function in terms of cosine, and the cosecant function in terms of sine. \(\sec ^{-1} \frac{1}{x}\) is equivalent to \(\cos^{-1}x\), and \(\csc^{-1}\frac{1}{x}\) is equivalent to \(\sin^{-1}x.\)
02

- Add the inverse functions

Now add \(\cos^{-1}x\) and \(\sin^{-1}x\). Remember that the definition of the cosine of an angle in a right triangle is the adjacent side divided by the hypotenuse, while the sine is the opposite side divided by the hypotenuse.
03

- Prove Using Triangle Properties

Let A be the angle such that \(\cos A = x\) and \(\sin A = \sqrt{1 - x^2}\). Therefore, \(\sin^{-1}\sqrt{1 - x^2} + \cos^{-1}x = A + \frac{\pi}{2} - A\), which simplifies to \(\frac{\pi}{2}\), demonstrating the proof.
04

- Summarizing the proof

After comparing this resulting value to the right side of the given identity, it can be seen that both sides are indeed equal, which confirms the give identity is correct.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inverse Trigonometric Functions
Inverse trigonometric functions, also known as arcfunctions, are the inverse operations of the trigonometric functions. They allow us to find the angle that corresponds to a given trigonometric ratio.
For instance, if the sine of an angle is a certain value, the inverse sine (denoted as \( \sin^{-1} \)) provides the measure of that angle. Similar inverse functions exist for cosine (\

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Most popular questions from this chapter

In Exercises 91 to \(95,\) verify the identity. $$\frac{\cos (x+h)-\cos x}{h}=\cos x \frac{(\cos h-1)}{h}-\sin x \frac{\sin h}{h}$$

Solve for \(y\) in terms of \(x\). $$2 x=\frac{1}{2} \sin ^{-1} 2 y$$

In Exercises 73 to \(88,\) verify the identity. $$\cos (\alpha-\beta)-\cos (\alpha+\beta)=2 \sin \alpha \sin \beta$$

In Exercises 91 to \(95,\) verify the identity. $$\frac{1-\sin x+\cos x}{1+\sin x+\cos x}=\frac{\cos x}{\sin x+1}$$

Solve each equation for exact solutions in the interval \(0 \leq x<2 \pi\) $$\sin x-\cos x=1$$
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