Write ${cos}^{4} x - {tan}^{2} x$ $cos^4x-tan^2x$ in terms of (a) ${cos}^{2} x$ $cos^2x$ and (b) ${sin}^{2} x$ $sin^2x$ ?

Question

1613 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-14T14:23:03

Please see below.

In terms of ${cos}^{2} x$ $cos^2x$

${cos}^{4} x - {tan}^{2} x$ $cos^4x-tan^2x$

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

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

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

In terms of ${sin}^{2} x$ $sin^2x$

${cos}^{4} x - {tan}^{2} x$ $cos^4x-tan^2x$

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

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

Write cos^4x-tan^2xcos4x−tan2xcos^4x-tan^2x in terms of (a) cos^2xcos2xcos^2x and (b) sin^2xsin2xsin^2x?