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How do you prove $\frac{{sec}^{2} x - 1}{{sec}^{2} x} = {sin}^{2} x$ $(sec^2 x-1)/sec^2x=sin^2x$ ?

1 Answer

May 3, 2016

see below

Explanation:

Left Side: $= \frac{{sec}^{2} x - 1}{{sec}^{2} x}$ $=(sec^2x-1)/sec^2x$

$= \frac{{sec}^{2} x}{{sec}^{2} x} - \frac{1}{{sec}^{2} x}$ $=sec^2x/sec^2x -1/sec^2x$

$= 1 - {cos}^{2} x$ $=1-cos^2x$

$= {sin}^{2} x$ $=sin^2x$

$=$ $=$ Right Side

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