Question

If \(\sec^{-1}x = \csc^{-1}y\), then \(\cos^{-1}(1/x) + \cos^{-1}(1/y)\) is equal to

• A. \(\pi\)
• B. \(\pi/4\)
• C. \(-\pi/2\)
• D. \(\pi/2\)

Hint:

\(\sin^{-1}\theta + \cos^{-1}\theta = \pi/2\) for \(\theta \in [-1, 1]\).

Answer 1

The Correct Option is D

To solve this problem, we need to find the value of \(\cos^{-1}\left(\frac{1}{x}\right) + \cos^{-1}\left(\frac{1}{y}\right)\) given that \(\sec^{-1}x = \csc^{-1}y\).

First, let's understand the given equation:

• \(\sec^{-1}x\) means \(\theta = \sec^{-1}x\), so, \(\sec\theta = x\).
• \(\csc^{-1}y\) means \(\theta_1 = \csc^{-1}y\), so, \(\csc\theta_1 = y\).
• Since \(\sec^{-1}x = \csc^{-1}y\), then \(\theta = \theta_1\).

From the properties of inverse trigonometric functions, we have:

• \(\sec\theta = x\) implies \(\cos\theta = \frac{1}{x}\).
• \(\csc\theta_1 = y\) implies \(\sin\theta_1 = \frac{1}{y}\).

Since \(\theta = \theta_1\), we can express both \(\cos\theta\) and \(\sin\theta\) in terms of \(x\) and \(y\):

• \(\cos\theta = \frac{1}{x}\), therefore, \(\theta = \cos^{-1}\left(\frac{1}{x}\right)\).
• \(\sin\theta = \frac{1}{y}\), therefore, \(\theta = \sin^{-1}\left(\frac{1}{y}\right)\).

Adding both inverse functions, we get:

\(\cos^{-1}\left(\frac{1}{x}\right) + \sin^{-1}\left(\frac{1}{y}\right) = \theta + \theta = \theta + \theta\)

But, by the identity \(\cos^{-1}(a) + \sin^{-1}(a) = \frac{\pi}{2}\), and knowing that \(\theta = \cos^{-1}\left(\frac{1}{x}\right) = \sin^{-1}\left(\frac{1}{y}\right)\), the total sum is:

\(\cos^{-1}\left(\frac{1}{x}\right) + \cos^{-1}\left(\frac{1}{y}\right) = \frac{\pi}{2}\)

Thus, the final answer is \(\pi/2\).