# How do you prove  tan^2x / (secx - 1) = secx + 1 ?

Jan 2, 2016

Hint : ${\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right) - 1$ and the difference of square rule to prove it. Step by step working is shown below.

#### Explanation:

To prove ${\tan}^{2} \frac{x}{\sec \left(x\right) - 1} = \sec \left(x\right) + 1$

Use the identity $1 + {\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right)$

We can rewrite this as

${\tan}^{2} \left(x\right) = {\sec}^{2} \left(x\right) - 1$

Now back to our problem

LHS
$= {\tan}^{2} \frac{x}{\sec \left(x\right) - 1}$

$= \frac{{\sec}^{2} \left(x\right) - 1}{\sec \left(x\right) - 1}$

Recall the difference of square rule

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

We need to apply that for ${\sec}^{2} \left(x\right) - 1$

$= \frac{\left(\sec \left(x\right) - 1\right) \left(\sec \left(x\right) + 1\right)}{\sec \left(x\right) - 1}$

$= \frac{\cancel{\sec \left(x\right) - 1} \left(\sec \left(x\right) + 1\right)}{\cancel{\sec \left(x\right) - 1}}$

$= \sec \left(x\right) + 1 =$RHS

Therefore, LHS = RHS thus proved.

Jan 2, 2016

Start by deciding on the more difficult side to work on. In this case, it's the left side. Recall the Pythagorean trigonometric identity, $\textcolor{red}{{\tan}^{2} x} = \textcolor{g r e e n}{{\sec}^{2} x - 1}$. Using this identity, replace $\textcolor{red}{{\tan}^{2} x}$ in the equation with $\textcolor{g r e e n}{{\sec}^{2} x - 1}$.

Left side:

$\frac{\textcolor{red}{{\tan}^{2} x}}{\sec x - 1}$

$\frac{\textcolor{g r e e n}{{\sec}^{2} x - 1}}{\sec x - 1}$

Since "${\sec}^{2} x - 1$" is a difference of squares, it can be broken down into $\textcolor{\mathmr{and} a n \ge}{\sec x + 1}$ and $\textcolor{b l u e}{\sec x - 1}$.

$\frac{\left(\textcolor{\mathmr{and} a n \ge}{\sec x + 1}\right) \left(\textcolor{b l u e}{\sec x - 1}\right)}{\sec x - 1}$

You will notice that "$\sec x - 1$" appears both in the numerator and denominator, so they can both be cancelled out.

$\frac{\left(\sec x + 1\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\left(\sec x - 1\right)}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\left(\sec x - 1\right)}}}}$

$\sec x + 1$

$\therefore$, LS$=$RS.