How do I prove this equation is an identity? Secx-Cosx/Secx=Sin²x

Question

4193 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-20T15:52:36

See below

#(secx-cosx)/secx#

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

#=sin^2x#, as required. #square#

Answer 2 · 2018-03-20T19:08:09

see below

start with #LHS#

#(secx-cosx)/secx--(1)#

now #secx=1/cosx#

#(1)rarr((1/cosx)-cosx)/(1/cosx)#

#=((1-cos^2x)/cancel(cosx))/(1/cancel(cosx))#

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

but

#cos^x+sin^2x=1#

#:.(2)=sin^2x=RHS#

as required