Mathematics — Standard (041) — CBSE Class 10 board question

Option (A) $\dfrac{3}{4}$

$4\sec\theta = 5 \Rightarrow \sec\theta = \dfrac{5}{4}$, so hypotenuse = 5, adjacent = 4, opposite = $\sqrt{25-16} = 3$.
Therefore, $\cot\theta = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{4}{3}$...

Wait — $\cot\theta = \dfrac{4}{3}$, which is option (D).

Answer: (D) $\dfrac{4}{3}$

$4\sec\theta = 5 \Rightarrow \sec\theta = \dfrac{5}{4} \Rightarrow \cos\theta = \dfrac{4}{5}$. Using Pythagoras, $\sin\theta = \dfrac{3}{5}$. Thus $\cot\theta = \dfrac{\cos\theta}{\sin\theta} = \dfrac{4/5}{3/5} = \dfrac{4}{3}$.

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• From $\sec\theta = 5/4$: hypotenuse = 5, base = 4, so perpendicular = $\sqrt{25-16}=3$.
• $\cot\theta = \text{base}/\text{perpendicular} = 4/3$.
• Common error: confusing $\cot$ with $\tan$ or misidentifying sides. Always draw a right triangle when a trig ratio is given.