Pruebalo #(sinx-cosx)^2=1-1/csc^2x#?

1 Answer
Oct 24, 2016

#(sinx-cosx)^2=1-1/csc(2x)#

Explanation:

The question appears to be in Spanish and I am answering in English. I hope answer will satisfy you.

#(sinx-cosx)^2# - using #(a-b)^2=a^2+b^2-2ab#, we get

= #sin^2x+cos^2x-2sinxcosx#

= #1-sin2x# - (as #sin^2x+cos^2x=1# and #sin2x=2sinxcosx#)

= #1-1/csc(2x)# - as #sinA=1/cscA#

Note #(sinx-cosx)^2=1-1/csc(2x)# and not #1-1/csc^2x# as

#1-1/csc^2x=1-sin^2x=cos^2x#