Question #f2c09
3 Answers
Jun 3, 2017
Recall the following identities:
#sin2x = sin(x+x) = sinxcosx + cosxsinx#
# = 2sinxcosx#
#cos2x = cos(x+x) = cosxcosx - sinxsinx#
#= cos^2x - sin^2x#
#sin^2x + cos^2x = 1#
This gives:
#(2(sinx + cosx))/(2sinxcosx + cos^2x - sin^2x + 1)#
#= (2(sinx + cosx))/(2sinxcosx + cos^2x - cancel(sin^2x) + cancel(sin^2x) + cos^2x)#
#= (cancel(2)(sinx + cosx))/(cancel(2)(sinxcosx + cos^2x))#
#= cancel(sinx + cosx)/(cosxcancel((sinx + cosx)))#
#= 1/cosx -= color(blue)(secx)#
Jun 3, 2017
Proved
Jun 3, 2017
look at picture
Explanation:
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