Prove #(sin^2x-tan^2x)/(1-sec^2x) =sin^2x#? The only way I see is to change the #1-sec^2x# to a negative tangent, and cancel out the tangent on top. But, I’m pretty sure you can’t do that, can I?

1 Answer
Mar 9, 2018

See explanation...

Explanation:

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

#color(white)((sin^2 x - tan^2 x)/(1-sec^2 x)) = (sin^2 x(1 - 1 / cos^2 x))/(1-sec^2 x)#

#color(white)((sin^2 x - tan^2 x)/(1-sec^2 x)) = (sin^2 x(1 - sec^2 x))/(1-sec^2 x)#

#color(white)((sin^2 x - tan^2 x)/(1-sec^2 x)) = sin^2 x#

Note that the domains of #tan x# and #sec x# both exclude #x = pi/2 + kpi# for any integer #k#.

In addition note that we require #1-sec^2 x != 0# and hence we also need to exclude #x = kpi# for any integer #k#.

So the identity we found excludes #x = (kpi)/2# for any integer #k#