Question #f7e61

Question

1780 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-19T14:02:44

See below

#tan^2 x -sin^2x# #(1)#

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

#= sin^2x(sec^2x - 1)#

#=sin^2xtan^2x# #(2)#

#therefore (1) = (2)# #sf(QED)#

Answer 2 · 2017-07-19T14:05:18

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

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

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

Use the identity:
#tan^2x+1=sec^2x#

#tan^2x-sin^2x=sin^2x(tan^2x)# [Proven]