How do you simplify #tan^2x - (cot^2x + 1)/cot^2x#?

2 Answers
Noah G · Geoff K.
Dec 4, 2016

Use the identity #tanx = sinx/cosx# and #cotx = 1/tanx = 1/(sinx/cosx) = cosx/sinx#.

So #tan^2x-(cot^2x+1)/cot^2x#

#=sin^2x/cos^2x- ((cos^2x/sin^2x + 1)/(cos^2x/sinx^2))#

#=sin^2x/cos^2x- ((cos^2x + sin^2x)/sin^2x)/(cos^2x/sin^2x)#

Now use #sin^2theta + cos^2theta = 1#.

#=sin^2x/cos^2x - (1/sin^2x xx sin^2x/cos^2x)#

#=sin^2x/cos^2x - 1/cos^2x#

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

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

#= -1#

Hopefully this helps!

Geoff K.
Dec 4, 2016

We can make use of the Pythagorean identities:
#sin^2x+cos^2x=1#
#"    "1"    "+cot^2x=csc^2x#
#tan^2x+"    "1"   "=sec^2x#

Explanation:

#tan^2x-color(green)(cot^2x+1)/color(navy)(cot^2x)=tan^2x-color(green)(csc^2x)/color(navy)(cot^2x)#
#color(white)(tan^2x-(cot^2x+1)/cot^2x)=tan^2x-color(green)(1//cancel(sin^2x))/color(navy)(cos^2x//cancel(sin^2x))#
#color(white)(tan^2x-(cot^2x+1)/cot^2x)=tan^2x-sec^2x#
#color(white)(tan^2x-(cot^2x+1)/cot^2x)=-1#

You can also convert all the trig functions to #sin#'s and #cos#'s if you are not sure what to do; it's sort of a catch-all. It will always work, but it may take a bit longer.

Related questions
See all questions in Introduction to Twelve Basic Functions
Impact of this question
1869 views around the world
You can reuse this answer
Creative Commons License