Question

If $secA=\frac{2}{\sqrt{3}},$ find the value of $\frac{tanA}{cosA}+\frac{1+sinA}{tanA}$

Answer 1

Given, $secA=\frac{2}{\sqrt{3}}$

In $right\Delta ABC$
Since $secA=\frac{2}{\sqrt{3}}$
AB = $\sqrt{3}$ and AC = 2
Now, by Pythagoras theorem
$A{C}^2=A{B}^2+B{C}^2$
$2^2=(=\sqrt{3}{)}^2=B{C}^2$
$4=3+B{C}^2$
$B{C}^2=4-3$
$B{C}^2=1$
$BC=1$
So, $tanA=\frac{1}{\sqrt{3}};cosA=\frac{\sqrt{3}}{2};sinA=\frac{1}{2}$
$\frac{tanA}{cosA}+\frac{1+sinA}{tanA}=\frac{\frac{1}{\sqrt{3}}}{\frac{\sqrt{3}}{2}}+\frac{1+\frac{1}{2}}{\frac{1}{\sqrt{3}}}$
$=\frac{2}{3}+\frac{\frac{3}{2}}{\frac{1}{\sqrt{3}}}=\frac{2}{3}+\frac{3\sqrt{3}}{2}=\frac{4+9\sqrt{3}}{6}$