To solve for \(\cosec A\) given that \(\cot A = \frac{2p}{p^2-1}\), and since \(A\) is an acute angle, we will use trigonometric identities and relationships.
- Start with the identity \(\cosec^2 A = 1 + \cot^2 A\).
- Given \(\cot A = \frac{2p}{p^2-1}\), calculate \(\cot^2 A = \left(\frac{2p}{p^2-1}\right)^2\).
- Compute: \(\cot^2 A = \frac{4p^2}{(p^2-1)^2}\).
- Substitute into the identity: \(\cosec^2 A = 1 + \frac{4p^2}{(p^2-1)^2}\).
- Convert the "1" into a fraction with a common denominator: \(\cosec^2 A = \frac{(p^2-1)^2 + 4p^2}{(p^2-1)^2}\).
- Expand \((p^2-1)^2\): \((p^2-1)^2 = p^4 - 2p^2 + 1\).
- So, \(\cosec^2 A = \frac{p^4 - 2p^2 + 1 + 4p^2}{(p^2-1)^2}\) simplifies to \(\cosec^2 A = \frac{p^4 + 2p^2 + 1}{(p^2-1)^2}\).
- Recognize the numerator as a perfect square: \(\cosec^2 A = \frac{(p^2+1)^2}{(p^2-1)^2}\).
- Taking the square root gives us \(\cosec A = \frac{p^2+1}{p^2-1}\) because \(A\) is acute, and cosecant is positive for acute angles.
Thus, the value of \(\cosec A\) is \(\frac{p^2 + 1}{p^2-1}\), which matches the correct answer option.
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