How do I prove that #sin^2x# + #tan^2xsin^2x# = #tan^2x# ?
2 Answers
Sep 20, 2015
Use
Explanation:
By definition:
#tan x = sin x / cos x#
and by Pythagoras:
#sin^2 x + cos^2 x = 1#
So:
#tan^2 x = sin^2 x / cos^2 x = sin^2 x / cos^2 x(cos^2 x + sin^2 x)#
#= (sin^2x cos^2x)/cos^2 x + (sin^2 x sin^2 x) / cos^2 x#
#= sin^2 x + sin^2 x tan^2 x#
Sep 20, 2015
Using the rules: