Question 262 marksVery Short AnswerSection B

  1. (B) 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)\dfrac{(1 + \sin \theta)(1 - \sin \theta)}{(1 + \cos \theta)(1 - \cos \theta)}.

Introduction to Trigonometry
Trigonometric identities

Official answer

The value of the expression equals cot⁡2θ=(78)2=4964\cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}. It is a clean rational number because the messy 113\sqrt{113} hypotenuse from a triangle method would have cancelled out anyway.

Answer keywords:cot θ7/8difference of squares1 − sin²θ = cos²θ1 − cos²θ = sin²θcot²θ49/64Pythagorean identity

Marking scheme

  1. 1.1 mark: simplifying numerator and denominator to (1−sin⁡2θ)/(1−cos⁡2θ)(1 - \sin^2\theta)/(1 - \cos^2\theta) and applying sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 to get cos⁡2θ/sin⁡2θ\cos^2\theta/\sin^2\theta.
  2. 2.1 mark: identifying the expression as cot⁡2θ\cot^2\theta and substituting cot⁡θ=78\cot \theta = \frac{7}{8} to obtain 4964\frac{49}{64}.
  3. 3.Full marks for the equivalent triangle method if it correctly reaches 4964\frac{49}{64}; deduct for leaving the answer as cot⁡2θ\cot^2\theta without substituting the value.

Hint

Recognise (1+x)(1−x)=1−x2(1+x)(1-x) = 1 - x^2; then use 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta and 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta to reduce the whole thing to cot⁡2θ\cot^2\theta.

Quick oral answer

I use (1+a)(1−a)=1−a2(1+a)(1-a) = 1 - a^2, so the fraction becomes (1−sin⁡2θ)/(1−cos⁡2θ)=cos⁡2θ/sin⁡2θ=cot⁡2θ(1 - \sin^2\theta)/(1 - \cos^2\theta) = \cos^2\theta/\sin^2\theta = \cot^2\theta; with cot⁡θ=78\cot \theta = \frac{7}{8}, that is (78)2=4964\left(\frac{7}{8}\right)^2 = \frac{49}{64}.

Common mistakes

  1. 1.Getting the ratio upside down as tan⁡2θ=6449\tan^2\theta = \frac{64}{49} by mixing up which of 1−sin⁡2θ1 - \sin^2\theta and 1−cos⁡2θ1 - \cos^2\theta equals cos⁡2θ\cos^2\theta versus sin⁡2θ\sin^2\theta.
  2. 2.Forgetting to square the given ratio and writing the answer as 78\frac{7}{8} instead of (78)2=4964\left(\frac{7}{8}\right)^2 = \frac{49}{64}.
  3. 3.Wasting time building a triangle with irrational hypotenuse 113\sqrt{113} and making arithmetic slips, instead of simplifying symbolically to cot⁡2θ\cot^2\theta first.

Analysis and explanation

Recognise the algebraic identity (1+a)(1−a)=1−a2(1 + a)(1 - a) = 1 - a^2 hidden in the expression before touching any numbers.


Concept

  • Numerator (1+sin⁡θ)(1−sin⁡θ)=1−sin⁡2θ(1 + \sin \theta)(1 - \sin \theta) = 1 - \sin^2\theta; denominator (1+cos⁡θ)(1−cos⁡θ)=1−cos⁡2θ(1 + \cos \theta)(1 - \cos \theta) = 1 - \cos^2\theta.
  • The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 converts these to cos²θ and sin²θ respectively.
  • The ratio cos⁡2θ/sin⁡2θ\cos^2\theta/\sin^2\theta is exactly cot⁡2θ\cot^2\theta, so the whole expression equals cot⁡2θ=(78)2=4964\cot^2\theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}.

Common mistakes

  • Trying to find θ or building a triangle (base 77, perpendicular 88, hypotenuse 113\sqrt{113}) — much longer and invites errors with an irrational hypotenuse.
  • Stopping at cot⁡2θ\cot^2\theta without substituting the given value 78\frac{7}{8}.

Real-world

  • Reducing a messy expression to a single known quantity before plugging in numbers mirrors how engineers simplify formulas before substituting measured values.

Interesting facts

The identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is a direct restatement of the Pythagoras theorem on a right triangle whose hypotenuse is 1, which is why it works for every angle.

The expression looks complicated but collapses to a single ratio — a reminder that in trigonometry, spotting the difference-of-squares pattern (1+a)(1−a)(1+a)(1-a) can turn a four-factor expression into a one-line answer.

Question details

Q26 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
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 5: Trigonometry — Introduction to Trigonometry: Applying the identities sin²θ + cos²θ = 1 to simplify expressions (CBSE Class 10 Mathematics)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why does the answer come out to cot⁡2θ\cot^2\theta and not cot⁡θ\cot \theta?

Because both numerator (cos⁡2θ\cos^2\theta) and denominator (sin⁡2θ\sin^2\theta) are squared quantities. Their ratio is (cos⁡θ/sin⁡θ)2=cot⁡2θ(\cos \theta/\sin \theta)^2 = \cot^2\theta, so you must square the given value: (7/8)2=4964(7/8)^2 = \frac{49}{64}.

Can I solve it by making a triangle?

Yes. With cot⁡θ=78\cot \theta = \frac{7}{8} take base 7, perpendicular 8, hypotenuse 49+64=113\sqrt{49+64} = \sqrt{113}. Then cos⁡2θ/sin⁡2θ=49113÷64113=4964\cos^2\theta/\sin^2\theta = \frac{49}{113} \div \frac{64}{113} = \frac{49}{64}. It works but is longer and error-prone, so the identity method is preferred.

Which identity is essential here?

The Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, rearranged as 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta and 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta.