How do you simplify $\frac{tan 2 x}{2 tan x} - \frac{{sec}^{2} x}{1 - {tan}^{2} x}$ $(tan2x)/(2tanx)-sec^2x/(1-tan^2x)$ ?

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Answer 1 · 2016-09-23T09:12:25

$\frac{tan 2 x}{2 tan x} - \frac{{sec}^{2} x}{1 - {tan}^{2} x}$ $(tan2x)/(2tanx)-sec^2x/(1-tan^2x)$

$= \frac{2 tan x}{(2 tan x) (1 - {tan}^{2} x)} - \frac{{sec}^{2} x}{1 - {tan}^{2} x}$ $=(2tanx)/((2tanx)(1-tan^2x))-sec^2x/(1-tan^2x)$

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

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

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

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

How do you simplify (tan2x)/(2tanx)-sec^2x/(1-tan^2x)tan2x2tanx−sec2x1−tan2x(tan2x)/(2tanx)-sec^2x/(1-tan^2x)?