Question

Prove that. $2sin^2 \frac{π}{6}+cosec^2 \frac{7π}{6}–cos^2 \frac{π}{3}=\frac{3}{2}$

Answer 1

L.H.S. = $2sin^2 \frac{π}{6}+cosec^2 \frac{7π}{6}–cos^2 \frac{π}{3}$

$=2(\frac{1}{2})^2+cosec^2({\pi}+\frac{\pi}{6})(\frac{1}{2})^2$

$2×\frac{1}{4}+(-cosec\frac{\pi}{6})^2(\frac{1}{4})$

$=\frac{1}{2}+(-2)^2(\frac{1}{4})$

$=\frac{1}{2}+\frac{4}{4}=\frac{1}{2}+1=\frac{3}{2}$

$=R.H.S$