simplify 1-tan^2x/1+tan^2x

Answer 1

(1+tan^2(x))/(1-tan^2(x))

= 1/(1-tan^2(x))/(1+tan^2(x))

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

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

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

= 1/cos(2x)

=sec(2x)

check the answer with a few basic angles, 0, 30, 45 degrees

(1+tan^2(0))/(1-tan^2(0)) = 1/1 = 1 = sec2(0) = 1

(1+tan^2(30))/(1-tan^2(30)) = (1+1/3)/(1-1/3) = (4/3)/(2/3) = 4/2 = 2 sec60 = 1/cos60 = 1/(1/2) = 2

(1+tan^2(45))/(1-tan^2(45)) = (1+1)/(1-1) = 2/0 = undefined = sec2(45) = sec90 = 1/cos90 = 1/0 = undefined

UNLESS you actually (but unlikely) really meant the problem as

1-tan^2(x)/1 + tan^2(x)

= 1 -(tan^2(x)/1) + tan^2(x)

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

= 1

That is how you've written in, but it's common to forget parentheses in fractions where numerator or denominator has multiple terms, so odds are you didn't really mean this 2nd interpretation, so odds are good, you want sec(2x) as the simplification answer, although you can't get more "simple" than 1.

Answer 2

To start this question, remember (A) the Pythagorean trig identity (sin²x + cos²x = 1), (B) tanx = sinx/cosx, and (C) cos(2x)=cos²x - sin²x. This means we can rewrite the expression in terms of sinx and cosx and simplify.

(1-tan²x)/(1+tan²x)

[1 - (sin²x) / (cos²x)] / [1 + (sin²x) / (cos²x)]

**Plugged in (B)**

[(cos²x - sin²x) / (cos²x)] / [((cos²x + sin²x) / (cos²x)]

**Combined fractions in denominator and numerator**

[cos(2x) / 1]•[cos²x / cos²x]

**Plugged in (C) in numerator, (A) in denominator, and rearranged fractions with cos²x

denominators by themselves**

cos(2x)

**Simplified expression**

So the simplified version is cos(2x).