Question 262 marksVery Short AnswerSection B

OR

If cot⁡θ=78\cot \theta = \frac{7}{8}, then find the value of (1+sin⁡θ)(1−sin⁡θ)(1+cos⁡θ)(1−cos⁡θ)\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}.

Introduction to Trigonometry
Trigonometric Identities

Official answer

The expression simplifies to cot⁡2θ=4964\cot^2\theta = \frac{49}{64}. Using (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2-b^2, the numerator (1+sin⁡θ)(1−sin⁡θ)=1−sin⁡2θ=cos⁡2θ(1+\sin\theta)(1-\sin\theta) = 1-\sin^2\theta = \cos^2\theta, and the denominator (1+cos⁡θ)(1−cos⁡θ)=1−cos⁡2θ=sin⁡2θ(1+\cos\theta)(1-\cos\theta) = 1-\cos^2\theta = \sin^2\theta. So the ratio equals cos⁡2θsin⁡2θ=cot⁡2θ\frac{\cos^2\theta}{\sin^2\theta} = \cot^2\theta. Given cot⁡θ=78\cot\theta = \frac{7}{8}, cot⁡2θ=4964\cot^2\theta = \frac{49}{64}.

Answer keywords:cot thetatrigonometric identitysin squared plus cos squared equals 1a squared minus b squaredcot squared thetasimplification

Marking scheme

  1. 1.1 mark: correctly simplifying numerator and denominator to cos⁡2θ\cos^2\theta and sin⁡2θ\sin^2\theta respectively (using a2−b2a^2-b^2 identity and sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1), reducing expression to cot⁡2θ\cot^2\theta.
  2. 2.1 mark: correctly substituting cot⁡θ=78\cot\theta = \frac{7}{8} and computing cot⁡2θ=4964\cot^2\theta = \frac{49}{64} as the final answer.

Hint

Use (1+sin⁡θ)(1−sin⁡θ)=1−sin⁡2θ=cos⁡2θ(1+\sin\theta)(1-\sin\theta) = 1 - \sin^2\theta = \cos^2\theta and (1+cos⁡θ)(1−cos⁡θ)=1−cos⁡2θ=sin⁡2θ(1+\cos\theta)(1-\cos\theta) = 1 - \cos^2\theta = \sin^2\theta; the expression reduces to cot⁡2θ\cot^2\theta.

Quick oral answer

I simplify the expression using (1+sin⁡θ)(1−sin⁡θ)=cos⁡2θ(1+\sin\theta)(1-\sin\theta) = \cos^2\theta and (1+cos⁡θ)(1−cos⁡θ)=sin⁡2θ(1+\cos\theta)(1-\cos\theta) = \sin^2\theta, so it becomes cot⁡2θ\cot^2\theta. Since cot⁡θ=78\cot\theta = \frac{7}{8}, cot⁡2θ\cot^2\theta = 49/64.

Common mistakes

  1. 1.Trying to find sin⁡θ\sin\theta and cos⁡θ\cos\theta individually first (using the 7-8-113\sqrt{113} triangle) instead of simplifying the expression algebraically — this makes the problem unnecessarily complicated and error-prone.
  2. 2.Forgetting the identity (1+x)(1−x)=1−x2(1+x)(1-x) = 1-x^2 and instead attempting to expand incorrectly.
  3. 3.Confusing cot⁡2θ\cot^2\theta with tan⁡2θ\tan^2\theta while substituting the given value, giving the reciprocal 6449\frac{64}{49} instead of 4964\frac{49}{64}.

Analysis and explanation

Simplify the expression algebraically to cot⁡2θ\cot^2\theta before substituting the given value — this avoids messy computation.


Concept

  • (a+b)(a−b)=a2−b2(a+b)(a-b) = a^2-b^2 gives (1+sin⁡θ)(1−sin⁡θ)=1−sin⁡2θ=cos⁡2θ(1+\sin\theta)(1-\sin\theta) = 1-\sin^2\theta = \cos^2\theta.
  • Similarly, (1+cos⁡θ)(1−cos⁡θ)=1−cos⁡2θ=sin⁡2θ(1+\cos\theta)(1-\cos\theta) = 1-\cos^2\theta = \sin^2\theta, so the ratio reduces to cot⁡2θ\cot^2\theta.

Key Points

  • Substitute cot⁡θ=78\cot\theta = \frac{7}{8} only after simplification to get cot⁡2θ\cot^2\theta = 4964\frac{49}{64}.

Common Mistakes

  • Trying to find sin⁡θ,cos⁡θ\sin\theta, \cos\theta individually first, which forces an ugly irrational hypotenuse (113\sqrt{113}).
  • Not recognising 1−sin⁡2θ1-\sin^2\theta and 1−cos⁡2θ1-\cos^2\theta as standard identity forms.

Exam Tip

  • Spotting patterns like (1−sin⁡2θ)(1-\sin^2\theta), (sec⁡2θ−tan⁡2θ)(\sec^2\theta-\tan^2\theta) and simplifying before substituting saves valuable exam time.

Interesting facts

The identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is essentially the Pythagorean theorem rewritten for a unit circle (hypotenuse = 1), connecting trigonometry directly to geometry.

This type of 'simplify before substituting' question is a favourite CBSE VSA pattern because it rewards conceptual understanding over brute-force calculation.

The three Pythagorean trigonometric identities (sin⁡2+cos⁡2=1\sin^2+\cos^2=1, 1+tan⁡2=sec⁡21+\tan^2=\sec^2, 1+cot⁡2=csc⁡21+\cot^2=\csc^2) were systematized in their modern algebraic form only in the 16th-17th century, even though the underlying geometric relationships were known to ancient Indian and Greek mathematicians.

Question details

Q26 of 56 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Standard)
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Introduction to Trigonometry
Topic
Trigonometric Identities
Bloom's level
apply
Syllabus topic
Unit V: Trigonometry — Chapter: Introduction to Trigonometry (Trigonometric Identities: sin²θ + cos²θ = 1)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why not directly compute sin⁡θ\sin\theta and cos⁡θ\cos\theta from cot⁡θ=78\cot\theta = \frac{7}{8}?

Because that would require a right triangle with hypotenuse 49+64=113\sqrt{49+64} = \sqrt{113}, an irrational number, making the calculation messy. Simplifying the expression to cot⁡2θ\cot^2\theta first avoids this altogether.

Is (1+sin⁡θ)(1−sin⁡θ)(1+\sin\theta)(1-\sin\theta) always equal to cos⁡2θ\cos^2\theta?

Yes, for any angle θ, this follows directly from the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, so 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta always holds.