Q1. [1] § 8.2 Trigonometric Ratios § 8.3 Trigonometric Ratios of Some Specific Angles § 8.5 Summary
In a right triangle ABC, right-angled at A, if $\sin B = \frac{1}{4}$, then the value of $\sec B$ is
- A 4
- B $\frac{\sqrt{15}}{4}$
- C $\sqrt{15}$
- D $\frac{4}{\sqrt{15}}$
Previously asked in CBSE board exam
2025 30/6/1 Q18
Generated by claude-sonnet-4-6 · 2026-06-15 10:26 · grounding rag
Model Answer
Option D: $\dfrac{4}{\sqrt{15}}$
Given $\sin B = \dfrac{1}{4}$, so opposite = 1, hypotenuse = 4. By Pythagoras theorem, adjacent $= \sqrt{4^2 - 1^2} = \sqrt{15}$. Therefore, $\sec B = \dfrac{\text{hypotenuse}}{\text{adjacent}} = \dfrac{4}{\sqrt{15}}$.
Source: Chapter 8, Section 8.2
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Explanation
- Examiners expect you to use the Pythagoras theorem to find the missing side, then apply the definition $\sec B = \dfrac{\text{hypotenuse}}{\text{adjacent side}}$.
- Key identity shortcut: $\sec^2 B = 1 + \tan^2 B$, or use $\sin^2 B + \cos^2 B = 1 \Rightarrow \cos B = \dfrac{\sqrt{15}}{4}$, so $\sec B = \dfrac{4}{\sqrt{15}}$.
- Don't confuse $\sec B$ with $\cos B$ — they are reciprocals of each other.
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