Question

If $5\tan A=\sqrt{7}$ and angle A lies in the $3^{rd}$ quadrant then $cosecA=?$

Answer 1

$5\tan A=\sqrt{7}$

$coseA=?$

$\tan A=\frac{\sqrt{7}}{5}$

In third quadrat cosec A is negative

$(hyp{)}^2=(\sqrt{7}x{)}^2+(5x{)}^2$

$=7x^2+5x^2$

$=13x^2$

hyp $=\sqrt{13}x$ where x is a common factor

$\therefore cosecA=\frac{hyp}{opp}=\frac{\sqrt{13}x}{\sqrt{7}x}=\frac{\sqrt{13}}{\sqrt{7}}$

$cosecA=-\frac{\sqrt{1}3}{\sqrt{7}}$