Question #7d3bd

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Answer 1 · 2016-11-17T19:37:31

see below

${cos}^{2} x + {cos}^{4} x = 1$ $cos^2x+cos^4x=1$

${cos}^{4} x = 1 - {cos}^{2} x = {sin}^{2} x$ $cos^4x=1-cos^2x=sin^2x$

Therefore

${tan}^{2} x + {tan}^{4} x = {tan}^{2} x (1 + {tan}^{2} x)$ $tan^2x+tan^4x=tan^2x(1+tan^2x)$

$= {tan}^{2} x {sec}^{2} x$ $=tan^2xsec^2x$

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

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

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

$= \frac{{sin}^{2} x}{{sin}^{2} x}$ $=sin^2x/sin^2x$

$= 1$ $=1$