If sinx=45 sin ⁡ x = 4 5 , find 1+cot2xcsc2x−1 1 + cot 2 ⁡ x csc 2 ⁡ x − 1 .

If $\sin x=\frac{4}{5}$, find $\frac{1+{\cot }^{2}x}{{\csc }^{2}x-1}$.

Answer Details

$\sin x=\frac{opp}{Hyp}=\frac{4}{5}$
5${}^2$ = 4${}^2$ + adj${}^2$
adj${}^2$ = 25 - 16 = 9
adj = $\sqrt{9}$ = 3
$\tan x=\frac{4}{3}$
$\cot x=\frac{1}{\frac{4}{3}}=\frac{3}{4}$
${\cot }^{2}x=(\frac{3}{4}{)}^2=\frac{9}{16}$
$\csc x=\frac{1}{\sin x}$
= $\frac{1}{\frac{4}{5}}=\frac{5}{4}$
${\csc }^{2}x=(\frac{5}{4}{)}^2=\frac{25}{16}$
$\therefore \frac{1+{\cot }^{2}x}{{\csc }^{2}x-1}=\frac{1+\frac{9}{16}}{\frac{25}{16}-1}$
= $\frac{25}{16}÷\frac{9}{16}$
= $\frac{25}{9}$