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

Question

Anna Allen · Accepted Answer

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!

Madelyn Anderson · Answer

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

Christian Antunez · Answer

undefined To simplify the expression tan^2(x) - (cot^2(x) + 1)/cot^2(x), we can use trigonometric identities.

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

Using the identity cot^2(x) = 1/tan^2(x), we can simplify further:

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

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

= -1