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

2 Answers
Jan 2, 2016

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

Explanation:

To prove tan^2(x)/(sec(x)-1) = sec(x)+1

Use the identity 1+tan^2(x) = sec^2(x)

We can rewrite this as

tan^2(x) = sec^2(x)-1

Now back to our problem

LHS
=tan^2(x)/(sec(x)-1)

=(sec^2(x)-1)/(sec(x)-1)

Recall the difference of square rule

a^2-b^2 = (a-b)(a+b)

We need to apply that for sec^2(x) -1

=((sec(x)-1)(sec(x)+1))/(sec(x)-1)

=(cancel(sec(x)-1)(sec(x)+1))/cancel(sec(x)-1)

=sec(x)+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, color(red)(tan^2x)=color(green)(sec^2x-1). Using this identity, replace color(red)(tan^2x) in the equation with color(green)(sec^2x-1).

Left side:

color(red)(tan^2x)/(secx-1)

(color(green)(sec^2x-1))/(secx-1)

Since "sec^2x-1" is a difference of squares, it can be broken down into color(orange)(secx+1) and color(blue)(secx-1).

((color(orange)(secx+1))(color(blue)(secx-1)))/(secx-1)

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

((secx+1)color(red)cancelcolor(black)((secx-1)))/color(red)cancelcolor(black)((secx-1))

secx+1

:., LS=RS.