Proving Trig Identity I'm trying to prove the following identity:
$$\frac{\tan^2(x)+1}{\csc^2(x)} = \tan^2(x)$$
I tried this page but I couldn't make any sense out of their steps listed.
 A: Recall that $\sec^2{x} = \tan^2{x} + 1$. So,
$$ \frac{\tan^2{x} + 1}{\csc^2{x}} = \frac{\sec^2{x}}{\csc^2{x}} = \frac{\sec^2{x}}{1} \cdot \frac{1}{\csc^2{x}} = \frac{1}{\cos^2{x}} \cdot \frac{\sin^2{x}}{1} = \frac{\sin^2{x}}{\cos^2{x}} = \tan^2{x}. $$
A: If you start with the identity $\tan^2 x + 1 = \sec^2 x$, you get $\frac{\sec^2 x}{\csc^2 x}$. Now rewrite that in terms of $\sin x$ and $\cos x$ and you'll get your result.
A: The way I'd choose is to simply convert everything into sines and cosines. So $\tan \rightarrow \sin/\cos$ and $\csc \rightarrow  1/\sin$:
$$\begin{align}
(\tan^2x +1)\frac1{\csc^2x}
&= \left( \frac{\sin^2x}{\cos^2x} + 1 \right)\sin^2 x \\
&= \left( \frac{\sin^2x + \cos^2x}{\cos^2x}  \right)\sin^2 x
\end{align}$$
hopefully you can see where to go from here?
A: Using $\csc(x)=\frac1{\sin(x)}$ and $\tan(x)=\frac{\sin(x)}{\cos(x)}$ you can write $$\frac{\tan^2(x)+1}{\csc^2(x)} = \sin^2(x)\cdot(\frac{\sin^2(x)}{cos^2(x)}+1) = \sin^2(x)\frac{\sin^2(x)+\cos^2(x)}{cos^2(x)}=\frac{\sin^2(x)\cdot1}{cos^2(x)}=\tan^2(x)$$
A: $$\frac{\tan^2(x)+1}{\csc^2(x)} = \tan^2(x)\\
({\tan^2(x)+1})\sin^2(x) = \tan^2(x)=\frac{\sin^2(x)}{\cos^2(x)}\\$$
Now, the RHS is $\frac{1}{\cos^2(x)}$, and the LHS is
$$(\frac{\sin^2(x)}{\cos^2(x)}+1)=(\frac{\sin^2(x)+\cos^2(x)}{\cos^2(x)})=\frac{1}{\cos^2(x)}$$
Both are the same. Q.E.D.
A: Another way , working the RHS $$ \frac{ \tan^2x + 1}{ \csc^2x} = \tan^2x $$ $$ \frac{ \tan^2x + 1}{ \csc^2x} = \tan^2x \frac{ \csc^2x}{ \csc^2x} $$ $$ \frac{ \tan^2x + 1}{ \csc^2x} = \frac{ \sec^2x}{ \csc^2x} $$
$$\frac{ \tan^2x + 1}{ \csc^2x} = \frac{ \tan^2x + 1}{ \csc^2x} $$
