Question: If $\pi = 180°$, verify that: $tan^2 \frac{\pi}{3} - cos^2\frac{\pi}{3} = \dfrac{sin^2 \frac{\pi}{3} - cos^4 \frac{\pi}{3}}{cos^2 \frac{\pi}{3}}$.

Solution:
We know,
$\pi = 180°$
To verify: $tan^2 \frac{\pi}{3} - cos^2\frac{\pi}{3} = \dfrac{sin^2 \frac{\pi}{3} - cos^4 \frac{\pi}{3}}{cos^2 \frac{\pi}{3}}$

Taking LHS
$= tan^2 \frac{\pi}{3} - cos^2\frac{\pi}{3}$

$= tan^2 \frac{180°}{3} - cos^2\frac{180°}{3}$

$= tan^2 60° - cos^2 60°$
$= (tan 60°)^2 - (cos 60°)^2$
$= (\sqrt{3})^2 - \left ( \dfrac{1}{2} \right )^2$

$= 3 - \dfrac{1}{4}$
$= \dfrac{12 -1}{4}$
$= \dfrac{11}{4}$

Taking RHS
$= \dfrac{sin^2 \frac{\pi}{3} - cos^4 \frac{\pi}{3}}{cos^2 \frac{\pi}{3}}$

$=  \dfrac{sin^2 \frac{180°}{3} - cos^4 \frac{180°}{3}}{cos^2 \frac{180°}{3}}$

$= \dfrac{sin^2 60°- cos^4 60° }{cos^2 60°}$

$= \dfrac{ (\frac{\sqrt{3}}{2} )^2 - (\frac{1}{2} )^4}{(\frac{1}{2} )^2}$

$= \dfrac{ \frac{3}{4} - \frac{1}{16}}{ \frac{1}{4}}$

$= \dfrac{ \frac{3×4 - 1}{16} }{ \frac{1}{4}}$

$= \dfrac{ 12-1}{16} × \dfrac{4}{1}$

$= \dfrac{11}{16} × 4$
$= \dfrac{11}{4}$
Since, LHS = $\dfrac{11}{4}$ = RHS

$tan^2 \frac{\pi}{3} - cos^2\frac{\pi}{3} = \dfrac{sin^2 \frac{\pi}{3} - cos^4 \frac{\pi}{3}}{cos^2 \frac{\pi}{3}}$

#verified

#SciPiPupil