Question

If $cosecA=\sqrt{2}$, find the value of $\frac{2si{n}^{2}A+3co{t}^{2}A}{4ta{n}^{2}A-co{s}^{2}A}.$

Answer 1

We have,
$cosecA=\sqrt{2}\Rightarrow \frac{1}{sinA}=\sqrt{2}\Rightarrow sinA=\frac{1}{\sqrt{2}}$

$Now,cosA=\sqrt{1-si{n}^{2}A}\Rightarrow cosA=\sqrt{1-{\left(\frac{1}{\sqrt{2}}\right)}^2}=\frac{1}{\sqrt{2}}$

$tanA=\frac{sinA}{cosA}\Rightarrow tanA=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=1$

And $cotA=\frac{1}{tanA}\Rightarrow cotA=\frac{1}{1}=1$

Hence,
$\frac{2si{n}^{2}A+3co{t}^{2}A}{4ta{n}^{2}A-co{s}^{2}A}=\frac{2\times {\left(\frac{1}{\sqrt{2}}\right)}^2+3(1{)}^2}{4(1{)}^2-{\left(\frac{1}{\sqrt{2}}\right)}^2}=\frac{2\times \frac{1}{2}+3}{4-\frac{1}{2}}=\frac{1+3}{7/2}=\frac{8}{7}$