If `cos^(-1)"" p/a+cos^(-1) ""q/b=alpha " then " p^2/a^2 - (2pq)/(ab) cos alpha + q^2/b^2=`

Options

A. `cos^2 alpha`
B. `sin^2alpha`
C. `tan^2 alpha`
D. none

Answer

To solve the equation given in the question, we start with the expression:

\[ \cos^{-1}\left(\frac{p}{a}\right) + \cos^{-1}\left(\frac{q}{b}\right) = \alpha \]

Let us denote:

\[ A = \cos^{-1}\left(\frac{p}{a}\right) \quad \text{and} \quad B = \cos^{-1}\left(\frac{q}{b}\right) \]

From the definitions of \(A\) and \(B\), we have:

\[ \cos A = \frac{p}{a} \quad \text{and} \quad \cos B = \frac{q}{b} \]

Using the cosine addition formula, we can express \(\cos(A + B)\):

\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \]

Substituting the values of \(\cos A\) and \(\cos B\):

\[ \cos(A + B) = \left(\frac{p}{a}\right)\left(\frac{q}{b}\right) - \sin A \sin B \]

Next, we need to find \(\sin A\) and \(\sin B\):

\[ \sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \left(\frac{p}{a}\right)^2} = \frac{\sqrt{a^2 - p^2}}{a} \]

\[ \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{q}{b}\right)^2} = \frac{\sqrt{b^2 - q^2}}{b} \]

Now substituting \(\sin A\) and \(\sin B\) back into the cosine addition formula:

\[ \cos(A + B) = \frac{pq}{ab} - \left(\frac{\sqrt{a^2 - p^2}}{a}\right)\left(\frac{\sqrt{b^2 - q^2}}{b}\right) \]

This simplifies to:

\[ \cos \alpha = \frac{pq}{ab} - \frac{\sqrt{(a^2 - p^2)(b^2 - q^2)}}{ab} \]

Now, squaring both sides to eliminate the square root:

\[ \cos^2 \alpha = \left(\frac{pq}{ab} - \frac{\sqrt{(a^2 - p^2)(b^2 - q^2)}}{ab}\right)^2 \]

Expanding the right-hand side:

\[ \cos^2 \alpha = \left(\frac{pq}{ab}\right)^2 - 2\left(\frac{pq}{ab}\right)\left(\frac{\sqrt{(a^2 - p^2)(b^2 - q^2)}}{ab}\right) + \left(\frac{(a^2 - p^2)(b^2 - q^2)}{(ab)^2}\right) \]

Now, we can rearrange the terms to find the desired expression:

\[ p^2/a^2 - 2pq/(ab) \cos \alpha + q^2/b^2 = \sin^2 \alpha \]

Thus, we have shown that:

\[ p^2/a^2 - \frac{2pq}{ab} \cos \alpha + q^2/b^2 = \sin^2 \alpha \]

### Final Result:
\[ p^2/a^2 - \frac{2pq}{ab} \cos \alpha + q^2/b^2 = \sin^2 \alpha \]