Is $\tan^2\theta+1=\large\frac{1}{\sin^2\theta}$ a Fundamental Identity Wrote this down during class, and I am wondering if I incorrectly transcribed from the board. Is this identity true? And if so, how? 
 A: It is not true. For example, let $\theta=\frac{\pi}{6}$ ($30$ degrees). Then $\tan^2\theta+1=\frac{1}{3}+1=\frac{4}{3}$ while $\frac{1}{\sin^2\theta}=4$. 
But $\tan^2\theta+1=\frac{1}{\cos^2\theta}$ is true.  
To show that the identity  $\tan^2\theta+1=\frac{1}{\cos^2\theta}$ holds, recall that $\sin^2\theta+\cos^2\theta=1$ and divide both sides by $\cos^2\theta$.
Maybe what was written on the board is $\cot^2\theta+1=\frac{1}{\sin^2\theta}$. This can be proved in a way very similar to the way we proved the identity $\tan^2\theta+1=\frac{1}{\cos^2\theta}$.
Remark: Are the identities $1+\tan^2\theta=\sec^2\theta$ and $1+\cot\theta=\csc^2\theta$  fundamental? I do not think they are. For one thing, they are too close relatives of the Pythagorean Identity $\cos^2\theta+\sin^2\theta=1$, which is more basic, and more generally useful. However, $1+\tan^2\theta=\sec^2\theta$ does come up fairly often, particularly when we are integrating trigonometric functions. And $\tan\theta$ does come up naturally in geometry, and then from $1+\tan^2\theta=\sec^2\theta$ the other trigonometric functions can be calculated.
A: To derived this quickly in your head, always note that
$$\cos^2(\theta) + \sin^2(\theta) = 1$$
from, here you could either divide the equation by $\cos^2(\theta)$ or $\sin^2(\theta)$,
Dividing by former, $\cos^2(\theta)$, we get,
$$1+tan^2(\theta) = \sec^2(\theta)$$
Dividing by the latter, $\sin^2(\theta)$, we get,
$$\cot^2(\theta) + 1 = \csc^2(\theta)$$
A: 
Above, we have a right triangle with an angle $\theta$, a hypotenuse (Hyp), an opposite side (Opp), and an adjacent side (Adj). By the Pythagorean Theorem we have
$$(Adj)^2+(Opp)^2=(Hyp)^2$$
Given the following definitions
$ \begin{array}{cc}
\sin\theta=\frac{Opp}{Hyp} & \csc\theta=\frac{Hyp}{Opp} \\
\cos\theta=\frac{Adj}{Hyp} & \sec\theta=\frac{Hyp}{Adj} \\
\tan\theta=\frac{Opp}{Adj} & \cot\theta=\frac{Adj}{Opp}\end{array}$
Now consider what would happen if we divided everything in the above (Pythagorean) equation by $(Hyp)^2$.
$$\frac{(Adj)^2+(Opp)^2}{(Hyp)^2}=\frac{(Hyp)^2}{(Hyp)^2}$$
$$\frac{(Adj)^2}{(Hyp)^2}+\frac{(Opp)^2}{(Hyp)^2}=\frac{(Hyp)^2}{(Hyp)^2}$$
$$\bigg(\frac{Adj}{Hyp}\bigg)^2+\bigg(\frac{Opp}{Hyp}\bigg)^2=\bigg(\frac{Hyp}{Hyp}\bigg)$$
$$\cos^2\theta+\sin^2\theta=1$$
As an exercise, instead of dividing by $(Hyp)^2$, try dividing by $(Adj)^2$ or by $(Opp)^2$ just to see what you come up with.
After that, try taking the identity (that you just derived)
$$\cos^2\theta+\sin^2\theta=1$$
and divide both sides by of the equation by $\sin^2\theta$. Then repeat the exercise but divide both side by $\cos^2\theta$ instead.
Lastly, try experimenting with the basic definitions and manipulate them.
For example, divide the numerator and denominator of the following identity by $Hyp$.
$$\tan\theta=\frac{Opp}{Adj}=\frac{Opp/Hyp}{Adj/Hyp}=\frac{?}{?}$$
Can you also express $\tan\theta$ in terms of $\csc\theta$ and $\sec\theta$?
The best way to approach math is to attack a concept from as many different angles as possible.
The worst thing that you can do is to write down equations from the black board, and try to memorize them without understanding where they come from.
