Question
Given: $\cos A = \frac{5}{13}$Evaluate: $(i)\frac{\sin A -\cot A }{2 \tan A };(ii)\cot A +\frac{1}{\cos A }$
✓
Answer
Consider the diagram below :
$\cos A =\frac{5}{13}$
$\text { i. e. } \frac{\text { base }}{\text { hypotenuse }}=\frac{5}{13} \Rightarrow \frac{ AB }{ AC }=\frac{5}{13}$
Therefore if length of $AB = 5x,$ length of $AC = 13x$
Since
$AB^2 + BC^2 = AC^2 \dots...[$ Using Pythagoras Theroem $]$
$(5x)^2 + BC^2 = (13x)^2$
$BC^2 = 169x^2 – 25x^2 = 144x^2$
$\therefore BC = 12x\dots ...($ perpendicular$)$
Now
$\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{12 x}{5 x}=\frac{12}{5}$
$\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{12 x}{13 x}=\frac{12}{13}$
$\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{5 x}{12 x}=\frac{5}{12}$
$\text { (i) } \frac{\sin A -\cot A }{2 \tan A }$
$=\frac{\frac{12}{13}-\frac{5}{12}}{2\left(\frac{12}{5}\right)}$
$=\frac{79}{156} \cdot \frac{5}{24}$
$=\frac{395}{3744}$
$\text { (ii) } \cot A +\frac{1}{\cos A }$
$=\frac{5}{12}+\frac{1}{\frac{5}{13}}$
$=\frac{5}{12}+\frac{13}{5}$
$=\frac{181}{60}$
$\cos A =\frac{5}{13}$
$\text { i. e. } \frac{\text { base }}{\text { hypotenuse }}=\frac{5}{13} \Rightarrow \frac{ AB }{ AC }=\frac{5}{13}$
Therefore if length of $AB = 5x,$ length of $AC = 13x$
Since
$AB^2 + BC^2 = AC^2 \dots...[$ Using Pythagoras Theroem $]$
$(5x)^2 + BC^2 = (13x)^2$
$BC^2 = 169x^2 – 25x^2 = 144x^2$
$\therefore BC = 12x\dots ...($ perpendicular$)$
Now
$\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{12 x}{5 x}=\frac{12}{5}$
$\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{12 x}{13 x}=\frac{12}{13}$
$\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{5 x}{12 x}=\frac{5}{12}$
$\text { (i) } \frac{\sin A -\cot A }{2 \tan A }$
$=\frac{\frac{12}{13}-\frac{5}{12}}{2\left(\frac{12}{5}\right)}$
$=\frac{79}{156} \cdot \frac{5}{24}$
$=\frac{395}{3744}$
$\text { (ii) } \cot A +\frac{1}{\cos A }$
$=\frac{5}{12}+\frac{1}{\frac{5}{13}}$
$=\frac{5}{12}+\frac{13}{5}$
$=\frac{181}{60}$
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