Prove this trigonometric identity:
$$\frac{1+\cos\theta}{\sec\theta-\tan\ \theta}+\frac{\cos\theta-1}{\sec\theta+\tan\ \theta}=2+2\tan\ \theta$$
I've simplified it until
$$\frac{2\cos^2\theta}{1-\sin\theta}$$
but couldn't get $2+2\tan\theta$ from it.
Best Answer
We have that
$$\frac{1+\cos\theta}{\sec\theta-\tan\theta}+\frac{\cos\theta-1}{\sec\theta+\tan\theta}=\frac{\cos\theta(1+\cos\theta)}{1-\sin\theta}+\frac{\cos\theta(\cos\theta-1)}{1+\sin\theta}=$$
$$=\frac{\cos\theta(1+\cos\theta)(1+\sin\theta)+\cos\theta(\cos\theta-1)(1-\sin\theta)}{1-\sin^2\theta}=$$
$$=\frac{(1+\cos\theta)(1+\sin\theta)+(\cos\theta-1)(1-\sin\theta)}{\cos\theta}=$$
$$=\frac{2\cos \theta+2\sin \theta}{\cos\theta}=2+2\tan\theta$$