Question #04652

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Answer 1 · 2017-02-07T22:10:23

see below

Left Hand Side:

$(tan^2x-1)/(cot^2x-1) = (sin^2x/cos^2x -1)/(cos^2x/sin^2x -1)$

$=(sin^2x/cos^2x -cos^2x/cos^2x)/(cos^2x/sin^2x -sin^2x/sin^2x)$

$=((sin^2x-cos^2x)/cos^2x)/((cos^2x -sin^2x)/sin^2x)$

$=(sin^2x-cos^2x)/cos^2x *sin^2x/(cos^2x -sin^2x)$

$=(sin^2x-cos^2x)/cos^2x *sin^2x/(-(sin^2x-cos^2x)$

$=cancel(sin^2x-cos^2x)/cos^2x *sin^2x/(-cancel(sin^2x-cos^2x)$

$=sin^2x/(-cos^2x)$

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

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

$=1/(-cos^2x)+cancel(cos^2x)/cancel(cos^2x)$

$=-1/cos^2x +1$

$=1-1/cos^2x$