Question
Given: $4\cot A = 3$;find :$(i)\sin A;(ii)\sec A;(iii)\operatorname{cosec}^2A - \cot^2A.$

Answer

Consider the diagram below :

$4 \cot A =3$
$\cot A =\frac{3}{4}$
i.e. $\frac{\text { base }}{\text { perpendicular }}=\frac{3}{4}$
$\Rightarrow \frac{ AB }{ BC }=\frac{3}{4}$
Therefore if length of $AB = 3x,$ length of $BC = 4x$
Since
$AB^2 + BC^2 = AC^2 \dots...[$ Using Pythagoras Theorem $]$
$(3x)^2 + (4x)^2 = AC^2$
$AC^2 = 9x^2 + 16x^2 = 25x^2$
$\therefore AC = 5x \dots...($ hypotenuse $)$
$(i)\sin A =\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{4 x}{5 x}=\frac{4}{5}$
$(ii)\sec A =\frac{\text { hypotenuse }}{\text { base }}=\frac{ AC }{ AB }=\frac{5 x}{3 x}=\frac{5}{3}$
$(iii) \operatorname{cosec} A =\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{ AC }{ BC }=\frac{5 x}{4 x}=\frac{5}{4}$
$\cot A =\frac{3}{4}$
$\operatorname{cosec}^2A – \cot^2 A$
$=\left(\frac{5}{4}\right)^2-\left(\frac{3}{4}\right)^2$
$=\frac{25-9}{16}$
$=\frac{16}{16}$
$= 1$

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