Correct Answer: $\frac{m^2-1}{m^2+1}$

Solution : $1 + \sin\theta = m\cos\theta$
⇒ $m=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}$
⇒ $m=\sec\theta+\tan\theta$
Squaring both sides, we get,
⇒ $m^2 = \sec^2\theta + \tan^2\theta + 2\sec\theta\tan\theta$
We know, $\sec^2\theta = 1+\tan^2\theta$
⇒ $m^2 = 1+\tan^2\theta + \tan^2\theta + 2\frac{\sin\theta}{\cos^2\theta}$
⇒ $m^2 = 1+2\frac{\sin^2\theta}{\cos^2\theta} +2\frac{\sin\theta}{\cos^2\theta}$
⇒ $m^2 = \frac{\cos^2\theta +2\sin^2\theta + 2\sin\theta}{\cos^2\theta}$
Applying componendo and dividendo rule,
⇒ $\frac{m^2-1}{m^2+1}=\frac{\cos^2\theta +2\sin^2\theta + 2\sin\theta -\cos^2\theta}{\cos^2\theta +2\sin^2\theta + 2\sin\theta+\cos^2\theta}$
⇒ $\frac{m^2-1}{m^2+1}=\frac{2\sin^2\theta+2\sin\theta}{2\cos^2\theta + 2\sin^2\theta +2\sin\theta}$
Using $\sin^2\theta + \cos^2\theta = 1$
⇒ $\frac{m^2-1}{m^2+1}=\frac{2\sin\theta(1+\sin\theta)}{2+2\sin\theta}$
⇒ $\frac{m^2-1}{m^2+1}=\frac{2\sin\theta(1+\sin\theta)}{2(1+\sin\theta)}$
⇒ $\sin\theta = \frac{m^2-1}{m^2+1}$
Hence, the correct answer is $\frac{m^2-1}{m^2+1}$.