To find the value of \cos\theta when \sin\theta = \frac{p^2 - 1}{p^2 + 1} and \theta is an acute angle, we can use the Pythagorean identity for sine and cosine:

\sin^2\theta + \cos^2\theta = 1

Given:

\sin\theta = \frac{p^2 - 1}{p^2 + 1}

First, find \sin^2\theta:

\sin^2\theta = \left(\frac{p^2 - 1}{p^2 + 1}\right)^2 = \frac{(p^2 - 1)^2}{(p^2 + 1)^2}

Express the identity:

\cos^2\theta = 1 - \sin^2\theta

Substitute:

\cos^2\theta = 1 - \frac{(p^2 - 1)^2}{(p^2 + 1)^2}

Simplify the expression:

\cos^2\theta = \frac{(p^2 + 1)^2 - (p^2 - 1)^2}{(p^2 + 1)^2}

Use the difference of squares formula:

\cos^2\theta = \frac{((p^2+1)+(p^2-1))((p^2+1)-(p^2-1))}{(p^2 + 1)^2}

Calculate the sums:

\cos^2\theta = \frac{(2p^2)(2)}{(p^2 + 1)^2} = \frac{4p^2}{(p^2 + 1)^2}

Now, find \cos\theta. Since \theta is an acute angle, \cos\theta is positive:

\cos\theta = \sqrt{\frac{4p^2}{(p^2 + 1)^2}} = \frac{2p}{p^2 + 1}

Therefore, the value of \cos\theta is:

Correct Answer: \frac{2p}{p^2 + 1}