Proving Identities Can you help me prove this identity: 
$$\frac{\sin x}{2\csc x}\left(\tan^2 x + \sin^2 x + \frac{\sin^2x}{\tan^2x}\right)= \frac{1}{2} \tan^2x$$
I have tried working on the left side and changing everything to sine and cosine, but I can't seem to get it down to only $\frac{1}{2} \tan^2 x$.
 A: Hint:: Work from the back leftward. We have $\frac{\sin^2 x}{\tan^2 x}=\cos^2 x$. Add $\sin^2 x$. We get $1$. Add $\tan^2 x$. We get $\sec^2 x$. Now it's almost over.  
A: Our identity that we need to prove:
$$\frac{\sin x}{2\csc x}\left(\tan^2 x + \sin^2 x +\frac{\sin^2 x}{\tan^2 x}\right)=\frac 12 \tan^2 x$$
Remember that $\dfrac{\sin^2 x}{\tan^2 x}=\cos^2 x$. This follows from the equality $\tan^2 x=\dfrac{\sin^2 x}{\cos^2 x}$.
$$\frac{\sin x}{2\csc x}\left(\tan^2 x+\sin^2 x+\cos^2 x\right)=\frac 12\tan^2 x$$
Remember that $\sin^2 x+\cos^2 x=1$
$$\frac{\sin x}{2\csc x}\left(\tan^2 x+1\right)=\frac 12\tan^2 x$$
Remember that $\tan^2 x+1=\sec^2 x$
$$\frac{\sin x}{2\csc x}\cdot \sec^2 x=\frac 12\tan^2 x$$
Remember that $\csc x =\dfrac{1}{\sin x}$ and that $\sec^2 x =\dfrac{1}{\cos^2 x}$.
$$\frac{\sin x}{\left(2\cdot\dfrac{1}{\sin x}\right)}\cdot \frac 1{\cos^2 x}=\frac 12\tan^2 x$$
$$\frac 12\cdot \sin x \cdot\sin x\cdot \frac 1{\cos^2 x}=\frac 12\tan^2 x$$
$$\frac 12\cdot \frac{\sin^2 x}{\cos^2 x}=\frac 12\tan^2 x$$
Finally, remember that $\tan^2 x=\dfrac{\sin^2 x}{\cos^2 x}$.
$$\frac 12\tan^2 x=\frac 12\tan^2 x \, \, \, \, \, \checkmark$$
$$\color{green}{\therefore \frac{\sin x}{2\csc x}\left(\tan^2 x + \sin^2 x +\frac{\sin^2 x}{\tan^2 x}\right)=\frac 12 \tan^2 x}$$
Hope I helped
