If \(\sec\theta + \tan\theta = p\), show that \(\sin\theta = \dfrac{p^2 - 1}{p^2 + 1}\).

Answer: Shown.

\(\sec^2\theta - \tan^2\theta = 1\) \(\therefore (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1\) \(\therefore \sec\theta - \tan\theta = \dfrac{1}{p}\)

Adding \(\sec\theta + \tan\theta = p\) and \(\sec\theta - \tan\theta = \dfrac{1}{p}\): \(2\sec\theta = p + \dfrac{1}{p} = \dfrac{p^2 + 1}{p}\) \(\therefore \sec\theta = \dfrac{p^2 + 1}{2p}\)

Subtracting: \(2\tan\theta = p - \dfrac{1}{p} = \dfrac{p^2 - 1}{p}\) \(\therefore \tan\theta = \dfrac{p^2 - 1}{2p}\)

\(\sin\theta = \dfrac{\sin\theta/\cos\theta}{1/\cos\theta} = \dfrac{\tan\theta}{\sec\theta} = \dfrac{p^2 - 1}{2p} \times \dfrac{2p}{p^2 + 1} = \dfrac{p^2 - 1}{p^2 + 1}\).