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

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