# How can this be reduced to the simplest form?

## $\frac{1 + {\tan}^{2} x}{1 + {\cot}^{2} x}$

Mar 6, 2018

See explanation

#### Explanation:

We want to simplify

$\frac{1 + {\tan}^{2} \left(x\right)}{1 + {\cot}^{2} \left(x\right)}$

Use the pythagorean trig identity

${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

$\implies 1 + {\cot}^{2} \left(x\right) = {\csc}^{2} \left(x\right) \textcolor{g r e e n}{\leftarrow \text{divided by} {\sin}^{2} \left(x\right)}$

$\implies 1 + {\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right) \textcolor{g r e e n}{\leftarrow \text{divided by} {\cos}^{2} \left(x\right)}$

Thus

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

Mar 6, 2018

${\tan}^{2} x$

#### Explanation:

$\text{using the "color(blue)"trigonometric identities}$

•color(white)(x)1+tan^2x=sec^2x

•color(white)(x)1+cot^2x=csc^2x

•color(white)(x)secx=1/cosx" and "cscx=1/sinx

$\Rightarrow \frac{1 + {\tan}^{2} x}{1 + {\cot}^{2} x}$

$= {\sec}^{2} \frac{x}{\csc} ^ 2 x$

$= \frac{\frac{1}{\cos} ^ 2 x}{\frac{1}{\sin} ^ 2 x}$

$= \frac{1}{\cos} ^ 2 x \times {\sin}^{2} x$

$= {\sin}^{2} \frac{x}{\cos} ^ 2 x$

$= {\tan}^{2} x$