WAEC Further Mathematics 2024 Paper

Evaluate: $\frac{\cos^2 300^{\circ} - 4\sin^2 120^{\circ}}{\tan^2 135^{\circ}}$

Cos 300º = 60º, Sin 120º = 60º, tan 135º = - 45º (using the idea of quadrant}, and the angles are all special.

cos 60º = $\frac{1}{2}$, sin 60º = $\frac{\sqrt{3}}{2}$, Tan(-45) = -1

$\frac{\cos^2 300^{\circ} - 4\sin^2 120^{\circ}}{\tan^2 135^{\circ}}$ = $\frac{\left(\frac{1}{2}\right)^2 - 4\left(\frac{\sqrt{3}}{2}\right)^2}{(-1)^2}$

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