If \(\sin x = \frac{4}{5}\), find \(\frac{1 + \cot^2 x}{\csc^2 x - 1}\).
A. \(\frac{13}{2}\)
B. \(\frac{25}{9}\)
C. \(\frac{3}{13}\)
D. \(\frac{4}{11}\)
Correct Answer: B
Explanation
\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\)
5\(^2\) = 4\(^2\) + adj\(^2\)
adj\(^2\) = 25 - 16 = 9
adj = \(\sqrt{9}\) = 3
\(\tan x = \frac{4}{3}\)
\(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\)
\(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\)
\(\csc x = \frac{1}{\sin x}\)
= \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\)
\(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\)
\(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\)
= \(\frac{25}{16} \div \frac{9}{16}\)
= \(\frac{25}{9}\)