1 / cos^4ф - 1 / cos^2ф = tan^4ф + tan^2ф
Right Hand Side $\displaystyle \, =\tan^4 \phi + \tan^2 \phi = \frac{\sin^4 \phi}{\cos^4 \phi} + \frac{\sin^2 \phi}{\cos^2 \phi}$.
Now note that:
$\displaystyle \frac{\sin^4 \phi}{\cos^4 \phi} = \frac{(\sin^2 \phi)^2}{\cos^4 \phi} = \frac{(1 - \cos^2 \phi)^2}{\cos^4 \phi} = \frac{1 - 2 \cos^2 \phi + \cos^4 \phi}{\cos^4 \phi} = \frac{1}{\cos^4 \phi} - \frac{2}{\cos^2 \phi} + 1$.
$\displaystyle \frac{\sin^2 \phi}{\cos^2 \phi} = \frac{1 - \cos^2 \phi}{\cos^2 \phi} = \frac{1}{\cos^2 \phi} - 1$.
Therefore the right hand side becomes .......
I did this one differently. I went like...
left side
step one multiply$\displaystyle \frac{1}{cos^2x}$ by $\displaystyle \frac{cos^2x}{cos^2x}$ and then add because we've achieved a common denominator.
Then using the pythagorean identity, we can change $\displaystyle 1-cos^2x $into $\displaystyle sin^2x$
Right side
convert the tan's into their proper sin and cosine values.
multiply $\displaystyle {sin^2x}{cos^2x}$ by $\displaystyle \frac{cos^2x}{cos^2x}$ and add because we've achieved a common denominator.
factor out a $\displaystyle sin^2x$
use the pythagorean identity to turn $\displaystyle sin^2x+cos^2x $ into 1
finally we end up with $\displaystyle \frac{sin^2x}{cos^4x}$
L=R