$\left[\frac{5}{8} \sec ^{2} 60^{\circ}- an ^{2} 60^{\circ}+\cos ^{2} 45^{\circ} ight]$ is equal t

Question

$\left[\frac{5}{8} \sec ^{2} 60^{\circ}-	an ^{2} 60^{\circ}+\cos ^{2} 45^{\circ}ight]$ is equal to\

Professor PC Thomas Classes · Answer

(c) :Consider, $\frac{5}{8} \sec ^{2} 60^{\circ}-	an ^{2} 60^{\circ}+\cos ^{2} 45^{\circ}$\ $=\frac{5}{8} 	imes(2)^{2}-(\sqrt{3})^{2}+\left(\frac{1}{\sqrt{2}}ight)^{2}=\frac{5}{8} 	imes 4-3+\frac{1}{2}=0$\