Understanding the Inverse Trigonometric Equation

We are given the equation $\sec^{-1} p - \csc^{-1} q = 0$, with the conditions that $p > 0$ and $q > 0$. Our goal is to find the value of $p^{-2} + q^{-2}$.

Let's start by rearranging the given equation:

$\sec^{-1} p = \csc^{-1} q$

Let's assume that this common value is $\theta$. So, we have:

$\sec^{-1} p = \theta \quad \text{and} \quad \csc^{-1} q = \theta$

Analyzing the Domain and Range

For the inverse secant function, $\sec^{-1} p$, the domain is $(-\infty, -1] \cup [1, \infty)$ and the principal value range is $[0, \pi/2) \cup (\pi/2, \pi]$.

For the inverse cosecant function, $\csc^{-1} q$, the domain is $(-\infty, -1] \cup [1, \infty)$ and the principal value range is $[-\pi/2, 0) \cup (0, \pi/2]$.

Given that $p > 0$ and $q > 0$, and considering the domains, we must have $p \ge 1$ and $q \ge 1$.

The common value $\theta$ must be in the intersection of the ranges where $\sec^{-1}$ and $\csc^{-1}$ are defined for positive arguments. This intersection is $(0, \pi/2]$.

If $\theta = \pi/2$, $\sec(\pi/2)$ is undefined. Therefore, $\theta$ cannot be $\pi/2$. This means $\theta$ must be in the interval $(0, \pi/2)$.

Using Trigonometric Definitions

From the inverse relations, if $\sec^{-1} p = \theta$, then by definition, $p = \sec \theta$.

Similarly, if $\csc^{-1} q = \theta$, then by definition, $q = \csc \theta$.

Now, we can express $\sec \theta$ and $\csc \theta$ in terms of basic trigonometric functions, sine and cosine:

• $p = \sec \theta = \frac{1}{\cos \theta}$
• $q = \csc \theta = \frac{1}{\sin \theta}$

From these relationships, we can find $\cos \theta$ and $\sin \theta$ in terms of $p$ and $q$:

• $\cos \theta = \frac{1}{p}$
• $\sin \theta = \frac{1}{q}$

Applying a Fundamental Trigonometric Identity

We know the fundamental trigonometric identity relating sine and cosine for any angle $\theta$:

$\sin^2 \theta + \cos^2 \theta = 1$

Now, substitute the expressions for $\sin \theta$ and $\cos \theta$ that we found:

$\left(\frac{1}{q}\right)^2 + \left(\frac{1}{p}\right)^2 = 1$

Squaring the terms gives:

$\frac{1}{q^2} + \frac{1}{p^2} = 1$

Recall that $x^{-n} = \frac{1}{x^n}$. So, we can write $\frac{1}{q^2}$ as $q^{-2}$ and $\frac{1}{p^2}$ as $p^{-2}$.

Therefore, the equation becomes:

$p^{-2} + q^{-2} = 1$

This is exactly the expression we were asked to find the value of.

Conclusion

The value of $p^{-2} + q^{-2}$ is 1.

Revision Table: Inverse Trigonometric Functions

Additional Information: Trigonometric Identities

Trigonometric identities are equations that are true for all values of the variable for which both sides of the equation are defined. The identity $\sin^2 \theta + \cos^2 \theta = 1$ is one of the most fundamental Pythagorean identities.

Other related Pythagorean identities derived from this one are:

• $1 + \tan^2 \theta = \sec^2 \theta$ (Divide $\sin^2 \theta + \cos^2 \theta = 1$ by $\cos^2 \theta$)
• $1 + \cot^2 \theta = \csc^2 \theta$ (Divide $\sin^2 \theta + \cos^2 \theta = 1$ by $\sin^2 \theta$)

These identities are crucial when simplifying trigonometric expressions or solving trigonometric equations, as seen in this problem where we used $\sin^2 \theta + \cos^2 \theta = 1$ to relate $p$ and $q$ derived from $\sec \theta$ and $\csc \theta$. Understanding the definitions and properties of inverse trigonometric functions and fundamental identities is key to solving such problems.