\frac{\sin \theta+\alpha}{\cos \theta-\alpha}=\frac{\sin\theta\cos\alpha+\cos\theta\sin\alpha}{cos\theta\cos\alpha +\sin\theta\sin\alpha}
.\qquad \qquad=\frac{\tan\theta+\tan\alpha}{1+\tan\theta\tan\alpha}=\frac{1-m}{1+m}
m=\frac{1+\tan\theta\tan\alpha-\tan\theta-\tan\alpha}{1+\tan\theta\tan\alpha+\tan\alpha+\tan\alpha}
.\qquad\qquad=\frac{(1-\tan\theta)(1-\tan\alpha)}{(1+\tan\theta)(1+\tan\theta)}=\tan(\pi/4-\theta)\tan(\pi/4-\alpha)
- Himanshu Giria In the second last step shouldn't it be tan a in 1 of the denominator
Upvote·0· Reply ·2013-08-08 20:34:50