If `sin^(-1) x = tan^(-1)y` what is the value of `(1)/(x^(2))-(1)/(y^(2))` ?

Options

A. 1
B. -1
C. 0
D. 2

Explanation

To solve the problem, we start with the given equation:

\[ \sin^{-1} x = \tan^{-1} y \]

Let's denote this common angle as \(\theta\):

\[ \theta = \sin^{-1} x \quad \text{and} \quad \theta = \tan^{-1} y \]

From these equations, we can express \(x\) and \(y\) in terms of \(\theta\):

1. From \(\theta = \sin^{-1} x\), we have:
\[ x = \sin \theta \]

2. From \(\theta = \tan^{-1} y\), we have:
\[ y = \tan \theta \]

Next, we need to find the value of:

\[ \frac{1}{x^2} - \frac{1}{y^2} \]

Substituting the expressions for \(x\) and \(y\):

\[ \frac{1}{x^2} = \frac{1}{(\sin \theta)^2} \quad \text{and} \quad \frac{1}{y^2} = \frac{1}{(\tan \theta)^2} \]

Thus, we can rewrite the expression as:

\[ \frac{1}{(\sin \theta)^2} - \frac{1}{(\tan \theta)^2} \]

Now, recall that:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]

Therefore,

\[ (\tan \theta)^2 = \frac{(\sin \theta)^2}{(\cos \theta)^2} \]

Substituting this into our expression gives:

\[ \frac{1}{(\sin \theta)^2} - \frac{1}{\frac{(\sin \theta)^2}{(\cos \theta)^2}} = \frac{1}{(\sin \theta)^2} - \frac{(\cos \theta)^2}{(\sin \theta)^2} \]

Now we can combine the fractions:

\[ \frac{1 - (\cos \theta)^2}{(\sin \theta)^2} \]

Using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\), we know that:

\[ 1 - (\cos \theta)^2 = \sin^2 \theta \]

Substituting this back into our expression yields:

\[ \frac{\sin^2 \theta}{(\sin \theta)^2} = 1 \]

Thus, the final answer is:

\[ \frac{1}{x^2} - \frac{1}{y^2} = 1 \]