Question

If sec A = $\sqrt{2}$, find : $\frac{3{\cot }^{2}A+2{\sin }^{2}A}{{\tan }^{2}A-{\cos }^{2}A}$

Answer 1

$\sec A=\sqrt{2}$

$\sec A=\sec {45}^{\circ }$

$A={45}^{\circ }$

Now, $\frac{3{\cot }^{2}A+2{\sin }^{2}A}{{\tan }^{2}A-{\cos }^{2}A}$

= $\frac{3{\cot }^{2}45+2{\sin }^{2}45}{{\tan }^{2}45-{\cos }^{2}45}$

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

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

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

= $8$