Prove the identity
$\displaystyle [1+(tanA)^2] / [1-(tanA)^2] = sec2A$
Thanks in advance
Yeah, I was working on the left side to get the right. Just found it easier to simplify the numerator first.
Yeah (you can derive this from $\displaystyle sin^2A+cos^2A=1$)
Convert your $\displaystyle tan^2 $ into sin and cosFor $\displaystyle [1 + (tanA)^2] / [1 - (tanA)^2] = sec2A$, starting with LHS,
$\displaystyle [1 + (tanA)^2] / [1 - (tanA)^2] = sec^2A / [1 - (tanA)^2] $
how do i continue?
$\displaystyle 1-tan^2A = 1 - \frac{sin^2A}{cos^2A}$
Then you'll want to make this into one fraction and since the LCM is $\displaystyle cos^2A$ we'll use that. $\displaystyle \left(1 = \frac{cos^2A}{cos^2A}\right)$
$\displaystyle \frac{cos^2A}{cos^2A} - \frac{sin^2A}{cos^2A}$
As these have the same denominator we can combine fractions so the above is equal to $\displaystyle \frac{cos^2A-sin^2A}{cos^2A}$
You should know that from your double angle identities that $\displaystyle cos^2A-sin^2A = cos(2A)$ and you should also know that when dividing by a fraction we flip it and multiply.
$\displaystyle \frac{sec^2A}{\frac{cos(2A)}{cos^2A}} = \frac{sec^2A}{cos(2A)} \times cos^2A$
From the definition of sec: $\displaystyle sec^2A \times cos^2A = 1$ so that cancels to leave $\displaystyle \frac{1}{cos(2A)}$ and again from the definition of sec:
$\displaystyle \frac{1}{cos(2A)} = sec(2A)$
Since this is equal to the RHS we've proved the identity is true