How do you verify #sec^2 x - cot^2 ( pi/2-x) =1#?

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Answer 1 · 2016-05-05T19:18:33

see below

Use Properties:
#cos(A-B)=cosAcosB+sinAsinB#

#sin(A-B)=sin A cos B-cos A sin B#

Left Side:#=sec^2x-cot^2(pi/2 -x)#

#=1/cos^2x - [(cos((pi/2)-x))/(sin((pi/2)-x))]^2#

#=1/cos^2x - [(cos(pi/2)cosx+sin(pi/2)sinx)/(sin(pi/2)cosx-cos(pi/2)sinx)]^2#

#=1/cos^2x - [(0*cosx+1*sinx)/(1*cosx-0*sinx)]^2#

#=1/cos^2x - [(sinx)/(cosx)]^2#

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

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

#=cos^2x/cos^2x#

#=1#

#=#Right Side