Question 91 markMCQSection A

Given that sin⁡θ=a/b\sin \theta = a/b, then cos θ is equal to :

(a) b/b2−a2b/\sqrt{b^2 - a^2} (b) b/ab/a (c) b2−a2/b\sqrt{b^2 - a^2}/b (d) a/b2−a2a/\sqrt{b^2 - a^2}

Introduction to Trigonometry
Trigonometric Ratios - Relation between sin and cos

Options

(A)b/b2−a2b/\sqrt{b^2 - a^2}
(B)b/ab/a
(C)b2−a2/b\sqrt{b^2 - a^2}/b
(D)a/b2−a2a/\sqrt{b^2 - a^2}

Official answer

(c) b2−a2/b\sqrt{b^2 - a^2}/b — with opposite = a and hypotenuse = b, adjacent=b2−a2\text{adjacent} = \sqrt{b^2 - a^2} by Pythagoras, so cos θ = adjacent/hypotenuse.

Answer keywords:trigonometric ratiossin thetacos thetaPythagoras theoremright triangleopposite adjacent hypotenuse

Marking scheme

  1. 1.1 mark: correct option (c) b2−a2/b\sqrt{b^2 - a^2}/b; full credit reasoning: opposite = a, hypotenuse = b, adjacent=b2−a2\text{adjacent} = \sqrt{b^2 - a^2} by Pythagoras, so cos θ = adjacent/hypotenuse.

Hint

Draw a right triangle with opposite = a, hypotenuse = b, find adjacent using Pythagoras, then cos θ = adjacent/hypotenuse.

Quick oral answer

If sin⁡θ=a/b\sin \theta = a/b, then in a right triangle opposite = a and hypotenuse = b, so adjacent=b2−a2\text{adjacent} = \sqrt{b^2 - a^2} by Pythagoras, giving cos⁡θ=b2−a2/b\cos \theta = \sqrt{b^2 - a^2}/b.

Common mistakes

  1. 1.Confusing cos θ with cosec θ (reciprocal of sin θ), giving answer b/ab/a.
  2. 2.Forgetting to apply the Pythagorean theorem to find the third side before writing the ratio.
  3. 3.Mixing up which side (opposite vs adjacent) goes in the numerator for cos⁡θ\cos \theta.

Analysis and explanation

Tests trigonometric ratios via a right-triangle picture rather than rote formula recall.


Concept

  • Draw a right triangle where sin θ = opposite/hypotenuse = a/ba/b, so opposite = a, hypotenuse = b.
  • By the Pythagorean theorem, adjacent = √(hypotenuse² − opposite²) = √(b² − a²).

Key points

  • cos θ = adjacent/hypotenuse = b2−a2/b\sqrt{b^2 - a^2}/b — option (c).

Common mistakes

  • Writing b/ab/a (that is cosec θ, the reciprocal of sin θ, not cos θ).
  • Mixing up opposite and adjacent, giving a/b2−a2a/\sqrt{b^2 - a^2} instead of the correct expression.
  • Trying to recall the formula from memory instead of sketching the triangle, which causes numerator/denominator swaps.

Interesting facts

The trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 (used implicitly here) was known in essentially modern form to Indian mathematician Bhaskaracharya (12th century) in his work Siddhanta Shiromani.

This type of 'given one ratio, find another' question is one of the most frequently repeated MCQ formats in CBSE Class 10 trigonometry across multiple years.

Question details

Q9 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Introduction to Trigonometry
Topic
Trigonometric Ratios - Relation between sin and cos
Bloom's level
apply
Syllabus topic
Unit 4: Trigonometry — Introduction to Trigonometry: Trigonometric Ratios and Identities
Author
CBSE
Reviewed by
Padhantu Academic Team

Spotted a mistake or something unclear?

Tell us and we will fix it.

Frequently asked questions

How do you find cos⁡θ\cos \theta if only sin⁡θ\sin \theta is given?

Use sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, so cos⁡θ=1−sin⁡2θ\cos \theta = \sqrt{1 - \sin^2\theta}, or use the right-triangle method: if sin θ = opposite/hypotenuse, find the adjacent side via Pythagoras and take cos θ = adjacent/hypotenuse.

What is the difference between cos⁡θ\cos \theta and cosec θ\text{cosec}\ \theta?

cos θ = adjacent/hypotenuse, while cosec θ = hypotenuse/opposite = 1/sin⁡θ1/\sin \theta; they are different ratios and should not be confused.