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

Assessment: JAMB UTME - Mathematics - 2019 Subject: General Mathematics

Question 1 Report

\frac{13}{2}

\frac{25}{9}

\frac{3}{13}

\frac{4}{11}

Answer Details

\sin x = \frac{o p p}{H y p} = \frac{4}{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}

∴ \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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