If `sectheta+tantheta=p ,` prove that `sintheta=(p^2-1)/(p^2+1)`

Explanation

`\frac{{p}^{2}-1}{{p}^{2}+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+(\tan ^{2} \theta+1)+2 \sec \theta \tan \theta} `

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

`=\frac{2 \tan 2}{2} \theta+2 \sec \theta \tan \theta `

`=\frac{2 \tan \theta(\tan \theta+\sec \theta)}{2 \sec \theta(\sec \theta+\tan \theta)} `

`=\frac{2 \tan \theta}{2 \sec \theta} `

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