Q9
1 markMCQSection A

  1. Given that sin⁡θ=ab\sin \theta = \frac{a}{b}, then cos θ is equal to : (a) bb2−a2\frac{b}{\sqrt{b^2-a^2}} (b) ba\frac{b}{a} (c) b2−a2b\frac{\sqrt{b^2-a^2}}{b} (d) ab2−a2\frac{a}{\sqrt{b^2-a^2}}

Introduction to Trigonometry
Trigonometric ratios

Options

(A)bb2−a2\frac{b}{\sqrt{b^2-a^2}}
(B)ba\frac{b}{a}
(C)b2−a2b\frac{\sqrt{b^2-a^2}}{b}
(D)ab2−a2\frac{a}{\sqrt{b^2-a^2}}
Official Answer

(c) b2−a2b\frac{\sqrt{b^2-a^2}}{b} — from cos⁡θ=1−sin⁡2θ=1−a2b2=b2−a2b\cos \theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{a^2}{b^2}} = \frac{\sqrt{b^2-a^2}}{b}.

trigonometric ratiossin cos identityPythagorean identitycos θ = √(b²−a²)/bright trianglehypotenuse b

Marking Scheme

  • 11 mark: correct option (c) b2−a2b\frac{\sqrt{b^2-a^2}}{b}.
  • 2Internal justification: cos⁡θ=1−a2b2=b2−a2b\cos \theta = \sqrt{1 - \frac{a^2}{b^2}} = \frac{\sqrt{b^2-a^2}}{b}.

Hint

Use cos⁡θ=1−sin⁡2θ\cos \theta = \sqrt{1 - \sin^2\theta}, or draw a right triangle with opposite a and hypotenuse b and find the adjacent side by Pythagoras.

Quick Oral Answer

From sin²θ plus cos²θ equals 1, cos θ equals the square root of 1 minus a squared over b squared, which simplifies to root b squared minus a squared, all over b.

Analysis & Explanation

Using sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 with sin⁡θ=ab\sin \theta = \frac{a}{b} gives cos⁡θ=b2−a2b\cos \theta = \frac{\sqrt{b^2 - a^2}}{b}, which also matches the right-triangle picture (opposite=a\text{opposite} = a, hypotenuse=b\text{hypotenuse} = b, adjacent=b2−a2\text{adjacent} = \sqrt{b^2 - a^2}).


Concept

  • Fundamental identity: cos⁡θ=1−sin⁡2θ\cos \theta = \sqrt{1 - \sin^2\theta} for acute θ\theta.
  • Right-triangle view: sin⁡θ=opposite/hypotenuse=ab⇒adjacent=b2−a2\sin \theta = \text{opposite}/\text{hypotenuse} = \frac{a}{b} \Rightarrow \text{adjacent} = \sqrt{b^2 - a^2} by Pythagoras.

Key steps

  • cos⁡θ=1−a2b2=b2−a2b2=b2−a2b\cos \theta = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{\frac{b^2 - a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b} → option (c).

Common mistakes

  • (a) bb2−a2\frac{b}{\sqrt{b^2-a^2}}: inverts numerator and denominator (looks like a secant-style flip).
  • (b) ba\frac{b}{a}: this is cosec θ, the reciprocal of sin θ, not cos θ.
  • (d) ab2−a2\frac{a}{\sqrt{b^2-a^2}}: this is tan θ (opposite/adjacent), not cos θ.

Real-world

  • Safe method: always keep the largest side (hypotenuse, here b) in the denominator for both sin and cos; only tan places a side other than the hypotenuse in the denominator.

Common Mistakes

  1. 1Flipping numerator and denominator to write bb2−a2\frac{b}{\sqrt{b^2-a^2}} (that is a secant-type expression, not cosine).
  2. 2Writing ba\frac{b}{a}, which is cosec θ (reciprocal of sin θ), by confusing 'reciprocal' with 'complementary' ratio.
  3. 3Giving ab2−a2\frac{a}{\sqrt{b^2-a^2}}, which is actually tan θ, by putting the adjacent side in the denominator instead of the hypotenuse.

Interesting Facts

The identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 is just the Pythagorean theorem applied to a right triangle with hypotenuse 1.

The word 'sine' comes from a mistranslation: the Sanskrit 'jya' passed through Arabic 'jiba' and was misread as 'jaib' (meaning fold or bay), giving Latin 'sinus'.

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Frequently Asked Questions

Why do we take only the positive square root for cos θ?

In Class 10 trigonometry θ is an acute angle (0° to 90°), where cosine is positive. So we take the positive root of 1−sin⁡2θ\sqrt{1 - \sin^2\theta}.

How does the right-triangle method give the same answer?

If sin⁡θ=ab\sin \theta = \frac{a}{b}, set opposite = a and hypotenuse = b. By Pythagoras, adjacent=b2−a2\text{adjacent} = \sqrt{b^2 - a^2}. Then cos⁡θ=adjacent/hypotenuse=b2−a2b\cos \theta = \text{adjacent}/\text{hypotenuse} = \frac{\sqrt{b^2 - a^2}}{b}.

What do the wrong options actually represent?

bb2−a2\frac{b}{\sqrt{b^2-a^2}} is sec-like, ba\frac{b}{a} is cosec θ, and ab2−a2\frac{a}{\sqrt{b^2-a^2}} is tan θ. Only b2−a2b\frac{\sqrt{b^2-a^2}}{b} is the cosine.