Question

If $4{\cos }^{2}A-3=0$ and $0^{\circ }\le A\le {90}^{\circ }$, where ${\tan }^{2}A+{\cos }^{2}A$ is $\frac{m}{12}$, then find $m.$

Answer 1

Angle $A$ is acute.

Given $4{\cos }^{2}A-3=0$

$\therefore {\cos }^{2}A=\frac{3}{4}$

We know that ${\sin }^{2}A=1-{\cos }^{2}A=1-\frac{3}{4}=\frac{1}{4}$

${\tan }^{2}A=\frac{{\sin }^{2}A}{{cos}^{2}A}=$ $\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

Then

${\tan }^2{A}^o+{\cos }^2{A}^o=\frac{1}{3}+\frac{3}{4}=\frac{13}{12}$

$\therefore m=13$