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

$\sin x = \frac{o p p}{H y p} = \frac{4}{5}$
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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Sine, Cosine And Tangent Of An Angle

Probability

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

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

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