Question
Given $: \tan A =\frac{4}{3}$, find $: \frac{\operatorname{cosecA}}{\cot A -\sec A }$
✓
Answer
Consider the diagram below :
$\tan A =\frac{4}{3}$
i.e. $\frac{\text { perpendicular }}{\text { base }}=\frac{4}{3}$
$\Rightarrow \frac{ BC }{ AB }=\frac{4}{3}$
Therefore if length of $AB = 3x,$ length of $BC = 4x$
Since
$AB^2 + BC^2 = AC^2 \dots... [$ Using Pythagoras Theorem $]$
$( 3x )^2 + (4x)^2 = AC^2$
$AC^2 = 9x^2 + 16x^2 = 25x^2$
$AC = 5x\dots ...($ hypotenuse $)$
Now
$\sec A =\frac{\text { hypotenuse }}{\text { base }}=\frac{ AC }{ AB }=\frac{5 x}{3 x}=\frac{5}{3}$
$\cot A =\frac{\text { base }}{\text { perpendicular }}=\frac{ AB }{ BC }=\frac{3 x}{4 x}=\frac{3}{4}$
$\operatorname{cosec} A =\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{ AC }{ BC }=\frac{5 x}{4 x}=\frac{5}{4}$
Therefore
$\frac{\cos e c A}{\cot A -\sec A }$
$=\frac{\frac{5}{4}}{\frac{3}{4}-\frac{5}{3}}$
$=\frac{\frac{5}{4}}{-\frac{11}{12}}$
$=-\frac{60}{44}$
$=-\frac{15}{11}$
$\tan A =\frac{4}{3}$
i.e. $\frac{\text { perpendicular }}{\text { base }}=\frac{4}{3}$
$\Rightarrow \frac{ BC }{ AB }=\frac{4}{3}$
Therefore if length of $AB = 3x,$ length of $BC = 4x$
Since
$AB^2 + BC^2 = AC^2 \dots... [$ Using Pythagoras Theorem $]$
$( 3x )^2 + (4x)^2 = AC^2$
$AC^2 = 9x^2 + 16x^2 = 25x^2$
$AC = 5x\dots ...($ hypotenuse $)$
Now
$\sec A =\frac{\text { hypotenuse }}{\text { base }}=\frac{ AC }{ AB }=\frac{5 x}{3 x}=\frac{5}{3}$
$\cot A =\frac{\text { base }}{\text { perpendicular }}=\frac{ AB }{ BC }=\frac{3 x}{4 x}=\frac{3}{4}$
$\operatorname{cosec} A =\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{ AC }{ BC }=\frac{5 x}{4 x}=\frac{5}{4}$
Therefore
$\frac{\cos e c A}{\cot A -\sec A }$
$=\frac{\frac{5}{4}}{\frac{3}{4}-\frac{5}{3}}$
$=\frac{\frac{5}{4}}{-\frac{11}{12}}$
$=-\frac{60}{44}$
$=-\frac{15}{11}$
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