Proving the identity $\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$ I am stuck with this trigonometric identity. It appeared in a question paper of mine, and I am wondering whether there is a print error or something, because I absolutely have no idea how to solve this. 

$$\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$$

I would really appreciate some inputs!
 A: Follow the usual process of manipulating only one side to make it look like the other side, and leave the other side completely alone.  The LHS is much more complicated, so let's start there and try to make it look like the RHS.
\begin{align*}
  \frac{\cos^2\theta + \tan^2\theta - 1}{\sin^2\theta}
    &= \frac{\cos^2\theta - 1 + \tan^2\theta}{\sin^2\theta}\\[0.3cm]
    &= \frac{\cos^2\theta - 1}{\sin^2\theta} + \frac{\tan^2\theta}{\sin^2\theta}\\[0.3cm]
    &= \frac{\cos^2\theta - 1}{1 - \cos^2\theta} + \tan^2\theta \cdot \frac1{\sin^2\theta}\\[0.3cm]
    &= \frac{-(1-\cos^2\theta)}{1 - \cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} \cdot \frac1{\sin^2\theta}\\[0.3cm]
    &= -1 + \frac1{\cos^2\theta}\\[0.3cm]
    &= -1 + \sec^2\theta\\[0.3cm]
    &= \tan^2\theta
\end{align*}
A: Hint:
Multiplying by $\sin^2 \theta$ and using the definition of $\tan$ and $1=\cos^2 \theta +\sin^2 \theta$,  we have:
$$
\cos^2 \theta +\frac{\sin^2 \theta}{\cos^2 \theta}-\cos^2 \theta -\sin^2 \theta = \sin^2 \theta \tan^2 \theta
$$
Now it is easy....
A: Multiply both sides with $\sin^2\theta$ and add the same to get
$$\cos^2\theta+\tan^2\theta-1+\sin^2\theta=\sin^2 \theta+ \sin^2 \theta\tan^2 \theta \implies\tan^2 \theta=\sin^2 \theta+ \sin^2 \theta\tan^2 \theta.$$
Now divide by $\tan^2 \theta$ to obtain
$$1=\cos^2 \theta+\sin^2 \theta,$$
which is true. 
A: $$\frac{\cos^2\theta + \frac{\sin^2 \theta}{\cos^2\theta}-1}{\sin^2\theta}=$$
$$= \frac{\cos^2 \theta +\frac{\sin^2\theta}{\cos^2 \theta}-\sin^2 \theta-\cos^2\theta}{\sin^2 \theta}$$
$$= \frac{\frac{\sin^2\theta-\cos^2\theta  \cdot \sin^2\theta}{\cos^2\theta}}{\sin^2\theta}$$
$$=\frac{\sin^2\theta \cdot (1-\cos^2\theta)}{\sin^2\theta \cdot \cos^2\theta}$$
$$=\frac{\sin^2\theta \cdot \sin^2 \theta}{sin^2\theta \cdot \cos^2\theta}$$
$$=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$$
A: $$\cos^2\theta=\dfrac1{1+\tan^2\theta}\text{ and }\sin^2\theta=\dfrac{\tan^2\theta}{1+\tan^2\theta}$$
Writing $\tan^2\theta=t$
$$\dfrac{\dfrac1{1+t}+t-1}{\dfrac t{1+t}}=\dfrac{(1+t^2-1)(1+t)}{(1+t)t}=t$$
A: Here,
$$L.H.S=\frac {cos^2\theta+Tan^2\theta-1}{sin^2\theta}$$
$$=\frac {Tan^2\theta - sin^2\theta}{sin^2\theta}$$
$$=\frac {Tan^2\theta}{sin^2\theta} - 1$$
$$=sec^2\theta -1$$
$$=Tan^2\theta=R.H.S$$
Proved.
