# How do you simplify csc^2(θ) / (1 + tan^2(θ) )?

Sep 21, 2015

${\cot}^{2} \theta$

#### Explanation:

Let's see, we have:

$\frac{{\csc}^{2} \theta}{1 + {\tan}^{2} \theta}$

In this case, I would probably start by rewriting everything in terms of sines and cosines, since we know that
$\tan \theta = \sin \frac{\theta}{\cos} \theta$
and
$\csc \theta = \frac{1}{\sin} \theta$

So we now have:

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

Here, I notice that the denominator can be simplified if I would multiply it by ${\cos}^{2} \theta$. Here is what I mean. The denominator is:
$1 + \frac{{\sin}^{2} \theta}{{\cos}^{2} \theta}$
and if I multiply ${\cos}^{2} \theta$, we get:
$\left(1 + \frac{{\sin}^{2} \theta}{{\cos}^{2} \theta}\right) \cdot {\cos}^{2} \theta$
Opening the parentheses, this becomes:

${\cos}^{2} \theta + {\sin}^{2} \theta$
which is our famous identity and is $= 1$.

Yes. The denominator becomes 1 if I would multiply by ${\cos}^{2} \theta$.
So let's do the following:

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

We can do this since any number over itself (except zero) is equal to 1.
So provided that $\theta \ne \frac{\pi}{2}$ or $- \frac{\pi}{2}$, we can multiply our equation by
${\cos}^{2} \frac{\theta}{\cos} ^ 2 \theta = 1$

Now, we know the denominator becomes 1, and we are left with just the "top":

$= {\cos}^{2} \frac{\theta}{\sin} ^ 2 \theta$
$= {\cot}^{2} \theta$

where $\cot \theta$ is the cotangent of theta.

Sep 22, 2015

Simplify: $f \left(x\right) = \frac{{\csc}^{2} x}{1 + {\tan}^{2} x}$

Ans: ${\cot}^{2} x$

#### Explanation:

Use the trig identity: $1 + {\tan}^{2} x = {\sec}^{2} x$. We get:

$f \left(x\right) = \frac{{\csc}^{2} x}{{\sec}^{2} x} = \frac{\frac{1}{\sin} ^ 2 x}{\frac{1}{\cos} ^ 2 x} =$
$= \frac{{\cos}^{2} x}{{\sin}^{2} x} = {\cot}^{2} x$