For an angle \(\theta\) with \(0^\circ<\theta<90^\circ\), it is given that \(\cot\theta=\displaystyle\frac{40}{9}\). Find \(\mathrm{cosec}\,\theta-\cot\theta\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{9}\)
\(\cot\theta\) is adjacent side ÷ opposite side in a right-angled triangle. Taking these sides as 40 and 9, Pythagoras gives the hypotenuse, and every other ratio follows. Since \(\theta\) is acute, all the ratios are positive.
Hypotenuse \(=\sqrt{40^2+9^2}=\sqrt{1600+81}=\sqrt{1681}=41\)
\(\mathrm{cosec}\,\theta=\displaystyle\frac{\text{hypotenuse}}{\text{opposite}}=\frac{41}{9}\)
\(\mathrm{cosec}\,\theta-\cot\theta=\displaystyle\frac{41}{9}-\frac{40}{9}=\frac{1}{9}\)