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

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