If $$\sec \theta + \tan \theta = p, 0^\circ < \theta < 90^\circ$$, then $$\frac{p^2 - 1}{p^2 + 1}$$ is equal to:

$$\sec \theta + \tan \theta = p$$

$$(\sec \theta + \tan \theta)^2 = p^2$$ ---(1)

$$\frac{p^2 - 1}{p^2 + 1}$$

From eq(1),

= $$\frac{(\sec \theta + \tan \theta)^2 - 1}{(\sec \theta + \tan \theta)^2 + 1}$$

= $$\frac{\sec^2 \theta + \tan^2 \theta + 2\sec \theta \tan \theta - 1}{\sec^2 \theta + \tan^2 \theta + 2\sec \theta \tan \theta + 1}$$

= $$\frac{(\sec^2 \theta - 1) + \tan^2 \theta + 2\sec \theta \tan \theta}{\sec^2 \theta + 2\sec \theta \tan \theta + (\tan^2 \theta + 1)}$$

= $$\frac{2 \tan^2 \theta + 2\sec \theta \tan \theta }{2\sec^2 \theta + 2\sec \theta \tan \theta}$$

= $$\frac{\tan^2 \theta + \sec \theta \tan \theta }{\sec^2 \theta + \sec \theta \tan \theta}$$

= $$\frac{\tan \theta(\tan \theta + \sec \theta)}{\sec \theta (\sec \theta + \tan \theta)}$$

= $$\frac{\tan \theta}{\sec \theta}$$ = $$\sin\theta$$