Question

Prove that $\frac{{\tan }^2{60}^{\circ }-2{\tan }^2{45}^{\circ }+{\sec }^2{30}^{\circ }}{3{\sin }^2{45}^{\circ }.{\sin }^2{90}^{\circ }+{\cos }^2{60}^{\circ }.{\cos }^2{0}^{\circ }}=\frac{4}{3}$

Answer 1

We know that

$\tan {60}^o=\sqrt{3},\tan {45}^o=1,\sec {30}^o=\frac{2}{\sqrt{3}}$

$\sin {45}^o=\frac{1}{\sqrt{2}},\sin {90}^o=1,\cos {60}^o=\frac{1}{2},cos{0}^o=1$

$\therefore \frac{{\tan }^2{60}^{\circ }-2{\tan }^2{45}^{\circ }+{\sec }^2{30}^{\circ }}{3{\sin }^2{45}^{\circ }\cdot {\sin }^2{90}^{\circ }+{\cos }^2{60}^{\circ }\cdot \cos 0^{\circ }}$

$=\frac{(\sqrt{3}{)}^2-2(1{)}^2+{\left(\frac{2}{\sqrt{3}}\right)}^2}{3{\left(\frac{1}{\sqrt{2}}\right)}^2(1{)}^2+{\left(\frac{1}{2}\right)}^2(1)}$

$=\frac{3-2+\frac{4}{3}}{3\left(\frac{1}{2}\right)\left(1\right)+\left(\frac{1}{4}\right)}$

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

$=\frac{\frac{3+4}{3}}{\frac{12+2}{8}}$

$=\frac{3+4}{3}\times \frac{8}{12+2}$

$=\frac{7}{3}\times \frac{8}{142}$

$=\frac{8}{3\times 2}=\frac{4}{3}$

Hence proved.