91Ó°ÊÓ

Let x = \(\cos ^{-1}\frac{p}{a}\) and y = $\cos ^{-1}\frac{q}{b} \(. Given that, x + y = \)\alpha$. Now, we have to find the value of: \(\frac{p^2}{a^2}-\frac{2pq}{ab} \cos \alpha + \frac{q^2}{b^2}\) which can be written as: \(\cos^2x - 2\cos{x}\cos{y}\cos(\cos^{-1}x + \cos^{-1}y) + \cos^2y\)

In this step, we will apply the cosine addition identity which is given by: \(\cos(x+y) = \cos{x}\cos{y} - \sin{x}\sin{y}\), to find the value of \(\cos(\cos^{-1}x + \cos^{-1}y)\) as we have x + y = \(\alpha\). So, we have: \(\cos \alpha = \cos{x}\cos{y} - \sin{x}\sin{y}\)

We have to find \(\sin{x}\) and \(\sin{y}\). We can use the Pythagorean identity for sine and cosine: \(\sin^{2}{x} + \cos^{2}{x} = 1\) So, \(\sin{x} = \sqrt{1 - \cos^2x} = \sqrt{1 - \left(\frac{p^2}{a^2}\right)} = \frac{\sqrt{a^2 - p^2}}{a}\) Similarly, \(\sin{y} = \sqrt{1 - \cos^2y} = \sqrt{1-\left(\frac{q^2}{b^2}\right)} = \frac{\sqrt{b^2 - q^2}}{b}\)

Now, we substitute these values of \(\sin{x}\), \(\sin{y}\), \(\cos{x}\) and \(\cos{y}\) in the expression of \(\cos \alpha\): \(\cos \alpha = \frac{p}{a} \cdot \frac{q}{b} - \frac{\sqrt{a^2 - p^2}}{a} \cdot \frac{\sqrt{b^2 - q^2}}{b}\) Now, let's replace this \(\cos \alpha\) value from the original expression: \(\cos^2x - 2\cos{x}\cos{y}\cos(\cos^{-1}x + \cos^{-1}y) + \cos^2y\) Insert the values of \(\cos{x}\), \(\cos{y}\) and \(\cos \alpha\): \(\frac{p^2}{a^2} - 2\left(\frac{p}{a}\cdot\frac{q}{b}\right) \left(\frac{p}{a}\cdot\frac{q}{b} - \frac{\sqrt{a^2 - p^2}}{a} \cdot \frac{\sqrt{b^2 - q^2}}{b}\right) + \frac{q^2}{b^2}\)

Simplify the given expression: \(\frac{p^2}{a^2} - \frac{2pq}{ab} \cdot \frac{p^2q^2}{a^2b^2} + \frac{2pq}{ab} \cdot \frac{\sqrt{a^2 - p^2}}{a}\cdot\frac{\sqrt{b^2-q^2}}{b} + \frac{q^2}{b^2}\) Cancelling common terms: \(\frac{p^2}{a^2} - \frac{p^3q^3}{a^3b^3} + \frac{2pq}{ab} \cdot \frac{\sqrt{a^2 - p^2}\sqrt{b^2-q^2}}{ab} + \frac{q^2}{b^2}\) Notice that this simplifies to: \(\frac{p^2}{a^2} + \frac{q^2}{b^2} - \frac{p^2q^2}{a^2b^2} \) Which is: \( \frac{p^2b^2}{a^2b^2} + \frac{q^2a^2}{a^2b^2} - \frac{p^2q^2}{a^2b^2}\) This can be written as: \( \frac{a^2q^2 + b^2p^2 - p^2q^2}{a^2b^2} = \frac{a^2q^2 + b^2p^2 - p^2q^2 + p^2q^2 - p^2q^2}{a^2b^2}\) This expression can be written as \( \frac{(a^2 - p^2)(b^2 - q^2)}{a^2b^2}\) Recall that we found earlier that \(\sin{x} = \frac{\sqrt{a^2 - p^2}}{a}\) and \(\sin{y} = \frac{\sqrt{b^2 - q^2}}{b}\) Thus, the expression becomes: \( \frac{(\sin^2x)(\sin^2y)}{(\sin{x}\sin{y})^2}\) Which simplifies to: \(\sin^2 \alpha\) Thus, the value of the given expression is \(\sin^2 \alpha\), which is option (a).