Question

Given: tan A = ${\displaystyle \frac{4}{3},\text{find}:\frac{\text{cosec}\text{A}}{\cot \text{A}–\sec \text{A}}}$

Answer 1

Consider the diagram below :

tan A = ${\displaystyle \frac{4}{3}}$

i.e. ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{4}{3}\Rightarrow \frac{\text{BC}}{\text{AB}}=\frac{4}{3}}$

Therefore if length of AB = 3x, length of BC = 4x

Since
AB² + BC² = AC²    ... [ Using Pythagoras Theorem ]

( 3x )² + (4x)² = AC²

AC² = 9x² + 16x² = 25x²

AC = 5x                   ...( hypotenuse )

Now

sec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=\frac{\text{AC}}{\text{AB}}=\frac{5x}{3x}=\frac{5}{3}}$

cot A = ${\displaystyle \frac{\text{base}}{\text{perpendicular}}=\frac{\text{AB}}{\text{BC}}=\frac{3x}{4x}=\frac{3}{4}}$

cosec A = ${\displaystyle \frac{\text{hypotenuse}}{\text{perpendicular}}=\frac{\text{AC}}{\text{BC}}=\frac{5x}{4x} =\frac{5}{4}}$

Therefore

${\displaystyle \frac{\text{cosec}\text{A}}{\cot \text{A}–\sec \text{A}}}$

= ${\displaystyle \frac{\frac{5}{4}}{\frac{3}{4} –\frac{5}{3}}}$

= ${\displaystyle \frac{\frac{5}{4}}{–\frac{11}{12}}}$

= ${\displaystyle –\frac{60}{44}}$

= ${\displaystyle –\frac{15}{11}}$