# Help required on solving Trigo sum..

• Dec 26th 2009, 04:48 AM
Punch
Help required on solving Trigo sum..
Prove the identity

$\displaystyle [1+(tanA)^2] / [1-(tanA)^2] = sec2A$

• Dec 26th 2009, 04:54 AM
e^(i*pi)
Quote:

Originally Posted by Punch
Prove the identity

[1+(tanA)^2] / [1-(tanA)^2] = sec2A

$\displaystyle 1+tan^2A = sec^2A$

$\displaystyle \frac{1}{cos^2A \cdot \left(\frac{cos^2A-sin^2A}{cos^2A}\right)}$

$\displaystyle cos^2A$ cancels leaving $\displaystyle \frac{1}{cos^2A-sin^2A} = \frac{1}{cos(2A)} = sec(2A) = RHS$
• Dec 26th 2009, 05:11 AM
Punch
Quote:

Originally Posted by e^(i*pi)

$\displaystyle 1+tan^2A = sec^2A$

$\displaystyle \frac{1}{cos^2A \cdot \left(\frac{cos^2A-sin^2A}{cos^2A}\right)}$

$\displaystyle cos^2A$ cancels leaving $\displaystyle \frac{1}{cos^2A-sin^2A} = \frac{1}{cos(2A)} = sec(2A) = RHS$

sorry but i only understood the part $\displaystyle 1+tan^2A = sec^2A$

this may provide a better picture, $\displaystyle [1+(tanA)^2] / [1-(tanA)^2] = sec2A$

it's a fraction
• Dec 26th 2009, 05:21 AM
Punch
so since $\displaystyle 1+tan^2A = sec^2A$

For $\displaystyle [1 + (tanA)^2] / [1 - (tanA)^2] = sec2A$, starting with LHS,

$\displaystyle [1 + (tanA)^2] / [1 - (tanA)^2] = sec^2A / [1 - (tanA)^2]$
how do i continue?
• Dec 26th 2009, 06:02 AM
skeeter
$\displaystyle \frac{1+\tan^2{A}}{1-\tan^2{A}} \cdot \frac{\cos^2{A}}{\cos^2{A}} = \frac{\cos^2{A} + \sin^2{A}}{\cos^2{A}-\sin^2{A}} = \frac{1}{\cos(2A)} = \sec(2A)$
• Dec 26th 2009, 06:07 AM
e^(i*pi)
Quote:

Originally Posted by Punch
sorry but i only understood the part $\displaystyle 1+tan^2A = sec^2A$

this may provide a better picture, $\displaystyle [1+(tanA)^2] / [1-(tanA)^2] = sec2A$

it's a fraction

Yeah, I was working on the left side to get the right. Just found it easier to simplify the numerator first.

Quote:

Originally Posted by Punch
so since $\displaystyle 1+tan^2A = sec^2A$

Yeah (you can derive this from $\displaystyle sin^2A+cos^2A=1$)

Quote:

For $\displaystyle [1 + (tanA)^2] / [1 - (tanA)^2] = sec2A$, starting with LHS,

$\displaystyle [1 + (tanA)^2] / [1 - (tanA)^2] = sec^2A / [1 - (tanA)^2]$
how do i continue?
Convert your $\displaystyle tan^2$ into sin and cos

$\displaystyle 1-tan^2A = 1 - \frac{sin^2A}{cos^2A}$

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