Question

Prove:

\[\frac{2(\tan x - \cot x)}{\tan^2 x - \cot^2 x} = \sin (2x)\]

Answer 1

convert tan and cot into sin and cos then use identities

Answer 2

Difference of squares says\[
\frac{2(\tan x - \cot x)}{\tan^2 x - \cot^2 x}=\frac{2(\tan x - \cot x)}{(\tan x - \cot x)(\tan x + \cot x)}=\frac{2}{(\tan x + \cot x)}
\]Dunno if that helps, I hate trig identities.

Answer 3

LS
\[=\frac{2(tanx-cotx)}{tan^2-cot^2x}\]\[=\frac{2(tanx-cotx)}{(tanx-cotx)(tanx+cotx)}\] \[=\frac{2}{(tanx+cotx)}\] \[=\frac{2}{\frac{sin^2x+cos^2x}{sinxcosx}}\] \[=\frac{2}{\frac{1}{sinxcosx}}\] \[=2sinxcosx\] \[=sin2x\]=RS

Answer 4

for the second line, I missed the x for tan^2 x :| Sorry!!!