Q26
2 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}.

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Marking Scheme

  • 11 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.
  • 21 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.

Analysis & 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.

Common Mistakes

  1. 1Trying 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. 2Forgetting the identity (1+x)(1−x)=1−x2(1+x)(1-x) = 1-x^2 and instead attempting to expand incorrectly.
  3. 3Confusing 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}.

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.

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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.