How do you verify ${sec}^{4} x - {sec}^{2} x = {tan}^{4} x + {tan}^{2} x$ $sec^4x-sec^2x=tan^4x+tan^2x$ ?

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Answer 1 · 2015-11-07T05:16:13

Verify ${sec}^{4} x - {sec}^{2} x = {tan}^{4} x + {tan}^{2} x$ $sec^4 x - sec^2 x = tan ^4 x + tan^2 x$

Left side $\to {sec}^{4} x - {sec}^{2} x$ $->sec^4 x - sec^2 x$

$= \frac{1}{{cos}^{4} x} - \frac{1}{{cos}^{2} x}$ $= 1/(cos^4 x) - 1/(cos^2 x)$

$= \frac{1 - {cos}^{2} x}{{cos}^{4} x}$ $= ( 1 - cos^2 x)/(cos^4 x)$

$= \frac{{sin}^{2} x}{{cos}^{4} x}$ $= sin^2 x/(cos^4 x)$

$= {tan}^{2} x (\frac{1}{{cos}^{2} x})$ $= tan^2 x(1/(cos^2 x))$

Apply the trig identity: $\frac{1}{{cos}^{2} x} = (1 + {tan}^{2} x)$ $1/(cos^2 x) = (1 + tan^2 x)$ , we get:

Left side $\to {tan}^{2} x (1 + {tan}^{2} x) = {tan}^{4} x + {tan}^{2} x .$ $-> tan^2 x(1 + tan^2 x)= tan^4 x + tan^2 x.$

How do you verify sec4x−sec2x=tan4x+tan2xsec^4x-sec^2x=tan^4x+tan^2x?