If \(\sin x = \frac{4}{5}\), find \(\frac{1 + \cot^2 x}{\csc^2 x – 1}\).

Explanation

\(\sin x = \frac{opp}{Hyp} = \frac{4}{5}\)

5\(^2\) = 4\(^2\) + adj\(^2\)

adj\(^2\) = 25 - 16 = 9

adj = \(\sqrt{9}\) = 3

\(\tan x = \frac{4}{3}\)

\(\cot x = \frac{1}{\frac{4}{3}} = \frac{3}{4}\)

\(\cot^2 x = (\frac{3}{4})^2 = \frac{9}{16}\)

\(\csc x = \frac{1}{\sin x}\)

= \(\frac{1}{\frac{4}{5}} = \frac{5}{4}\)

\(\csc^2 x = (\frac{5}{4})^2 = \frac{25}{16}\)

\(\therefore \frac{1 + \cot^2 x}{\csc^2 x - 1} = \frac{1 + \frac{9}{16}}{\frac{25}{16} - 1}\)

= \(\frac{25}{16} \div \frac{9}{16}\)

= \(\frac{25}{9}\)

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