How do you simplify ${(sec (x))}^{2} - 1$ $(sec(x))^2−1$ ?

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Answer 1 · 2015-09-07T08:00:14

Using the Pythagorean identity:

${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x = sec^2x - 1$

This is an application of the Pythagorean identities, namely:

$1 + {tan}^{2} x = {sec}^{2} x$ $1 + tan^2x = sec^2x$

This can be derived from the standard Pythagorean identity by dividing everything by ${cos}^{2} x$ $cos^2x$ , like so:

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2x = 1$

$\frac{{cos}^{2} x}{{cos}^{2} x} + \frac{{sin}^{2} x}{{cos}^{2} x} = \frac{1}{{cos}^{2} x}$ $cos^2x/cos^2x + sin^2x/cos^2x = 1/cos^2x$

$1 + {tan}^{2} x = {sec}^{2} x$ $1 + tan^2x = sec^2x$

From this identity, we can rearrange the terms to arrive at the answer to your question.

${tan}^{2} x = {sec}^{2} x - 1$ $tan^2x = sec^2x - 1$

It would help you in the future to know all three versions of the Pythagorean identities:

${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2x = 1$

$1 + {tan}^{2} x = {sec}^{2} x$ $1 + tan^2x = sec^2x$ (divide all terms by ${cos}^{2} x$ $cos^2x$ )

${cot}^{2} x + 1 = {csc}^{2} x$ $cot^2x + 1 = csc^2x$ (divide all terms by ${sin}^{2} x$ $sin^2x$ )

If you forget these, just remember how to derive them: by dividing by either ${cos}^{2} x$ $cos^2x$ or ${sin}^{2} x$ $sin^2x$ .

How do you simplify (sec(x))^2−1(sec(x))2−1(sec(x))^2−1?