Question

In Exercises 129 and \(130,\) verify each identity (a) \(\operatorname{arccsc} x=\arcsin \frac{1}{x}, \quad|x| \geq 1\) (b) \(\arctan x+\arctan \frac{1}{x}=\frac{\pi}{2}, \quad x>0\)

Short answer

(a) Given that \( \csc(\operatorname{arccsc} x) = x \) and \( \sin(\arcsin y) = y \), we can show that \( \operatorname{arccsc} x=\arcsin \frac{1}{x}, \quad|x| \geq 1 \).\n(b) If we let \( x = \tan A \), then \( x = \tan A \) and \( \frac{1}{x} = \tan(\frac{\pi}{2} - A) \), so \(\arctan x + \arctan \frac{1}{x} = A + \frac{\pi}{2} - A = \frac{\pi}{2} \).

Step-by-step solution

Part a: Relating arccsc x and arcsin 1/x

First, recall the definitions of arccsc (arc-cosecant) and arcsin (arc-sine). The function arc-cosecant is the inverse of cosecant, and arcsine is the inverse of sine. So \( \csc(\operatorname{arccsc} x) = x \) and \( \sin(\arcsin y) = y \). Using the identity \( \csc \theta = \frac{1}{\sin \theta} \), we can write \( x = \frac{1}{\sin(\arcsin \frac{1}{x})} \). As x equals \(\csc(\operatorname{arccsc} x)\), we found that \( \operatorname{arccsc} x=\arcsin \frac{1}{x}, \quad|x| \geq 1 \)

Part b: Verifying the identity \(\arctan x+\arctan \frac{1}{x}=\frac{\pi}{2}, \quad x>0\)

Let \( x = \tan A \) where 0 \< A \< \( \frac{\pi}{2} \). Then \(\arctan x = A \).Now, \( \frac{1}{x} \) will be \( \tan(\frac{\pi}{2} - A) \) since the tangent of a complementary angle is equal to the cotangent of the angle. Thus \( \arctan \frac{1}{x} = \frac{\pi}{2} - A \).Adding both equations: \( A + \frac{\pi}{2} - A = \frac{\pi}{2} \). Therefore, the identity is verified.