Prove that $\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$

Question

Question: $$\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}=\frac{(1-\cos\theta)\csc^2\theta}{\csc\theta+1}$$

Prove that L.H.S.=R.H.S.

My Efforts:

L.H.S.$$=\frac{\tan^2\theta(\csc\theta-1)}{1+\cos\theta}\times\frac{1-\cos\theta}{1-\cos\theta}\times\frac{\csc\theta+1}{\csc\theta+1}$$

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

Answer 1

Hint:

$$\tan^2\theta (\csc^2\theta-1) = \frac{\sin^2\theta}{\cos^2\theta}\left(\frac{1}{\sin^2\theta}-1\right) = \frac{\sin^2\theta}{\cos^2\theta}\left(\frac{1}{\sin^2\theta}-\frac{\sin^2\theta}{\sin^2\theta}\right) = \frac{\sin^2\theta}{\cos^2\theta}\left(\frac{1-\sin^2\theta}{\sin^2\theta}\right) = \frac{\sin^2\theta}{\cos^2\theta}\frac{\cos^2\theta}{\sin^2\theta} = 1$$

Answer 2

You are almost there.

Hint: $$\sin^2\theta+\cos^2\theta=1\implies \left(\frac{1}{\sin^2\theta}\right)(\sin^2\theta+\cos^2\theta)=\frac{1}{\sin^2\theta}$$

Also note that in the calculation you have reproduced, there is an error in the final denominator. It should be $(\csc\theta +1)$.

Answer 3

$$\tan^2\theta(\csc^2\theta-1)=(1-\cos^2\theta)\csc^2\theta,$$ $$\frac{\sin^2\theta}{\cos^2\theta}(\frac1{\sin^2\theta}-1)=(1-\cos^2\theta)\frac1{\sin^2\theta},$$ $$\frac{\sin^2\theta}{\cos^2\theta}\frac{\cos^2\theta}{\sin^2\theta}=\frac{\sin^2\theta}{\sin^2\theta}.$$