Question

(1 point) If tan x - -1/3, cosx > 0,, then sin 2x- cos 2x - tan 2x -

Answer 1

$\\ \sin 2x = \frac{2\tan x}{1+\tan^2x}=\frac{2\left ( \frac{-1}{3} \right )}{1+\left ( \frac{-1}{3} \right )^2}=\frac{\frac{-2}{3}}{1+\frac{1}{9}}=\frac{\frac{-2}{3}}{\frac{10}{9}}=\frac{-3}{5}\\ \\ \\ \cos 2x = \frac{1-\tan^2x}{1+\tan^2x}=\frac{1-\left ( \frac{-1}{3} \right )^2}{1+\left ( \frac{-1}{3} \right )^2}=\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{\frac{8}{9}}{\frac{10}{9}}=\frac{4}{5}\\ \\\\ \tan 2x = \frac{2\tan x}{1-\tan^2x}=\frac{2\left ( \frac{-1}{3} \right )}{1-\left ( \frac{-1}{3} \right )^2}=\frac{\frac{-2}{3}}{1-\frac{1}{9}}=\frac{\frac{-2}{3}}{\frac{8}{9}}=\frac{-3}{4}\\ \\\\ \\ \\ \cos 2\theta = 1-2\sin^2\theta =1-\left ( \frac{13}{85} \right )^2=1-\frac{169}{7225}=\frac{7056}{7225}$