Hello, rubeio!
\(\displaystyle \L\frac{1\,-\,\cot^2x}{1\,+\,\cot^2x}\)
+2⋅cos2x=1
I usually don't recommend changin everything to sines and cosines; there's usually a better way.
But this time, it works out nicely . . .
We have: \(\displaystyle \L\,\frac{1\,-\,\frac{\cos^2x}{\sin^2x}}{1\,+\,\frac{\cos^2x}{\sin^2x}}\)
+2⋅cos2x
Multiply top and bottom of the fraction by
sin2x:
\(\displaystyle \L\;\;\frac{\sin^2x\left(1\,-\,\frac{\cos^2x}{\sin^2x}\right)}{\sin^2x\left(1\,+\,\frac{\cos^2x}{\sin^2x}\right)}\)\(\displaystyle \,+\,2\cdot\cos^2x \;= \;\L\frac{\sin^2x\,-\,\cos^2x}{\sin^2x\,+\,\cos^2x}\)
+2cos2x
Since
sin2x+cos2x=1, we have:
sin2x−cos2x+2⋅cos2x=sin2x+cos2x=1
\(\displaystyle \L\frac{\sin^2x\,-\,\tan x}{\cos^2x\,-\,\cot x}\)
=tan2x
There are several approaches to this one . . . none of them are pleasant.
We have: \(\displaystyle \L\,\frac{\sin^2x\,-\,\frac{\sin x}{\cos x}}{\cos^2x\,-\,\frac{\cos x}{\sin x}}\)
Factor: \(\displaystyle \L\,\frac{\sin x\left(\sin x\,-\,\frac{1}{\cos x}\right)}{\cos x\left(\cos x\,-\,\frac{1}{\sin x}\right)} \;= \;\frac{\sin x\left(\frac{\sout{\sin x\cdot\cos x\,-\,1}}{\cos x}\right)}{\cos x\left(\frac{\sout{\sin x\cdot\cos x\,-\,1}}{\sin x}\right)} \;= \;\frac{\sin x\cdot\frac{1}{\cos x}}{\cos x\cdot\frac{1}{\sin x}}\)
\(\displaystyle \L\;\;\;=\;\frac{\sin^2x}{\cos^2x}\;=\;\left(\frac{\sin x}{\cos x}\right)^2\)
=tan2x