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If sin θ = ab\dfrac{a}{b}, then cos θ is equal to

  1. bb2−a2\dfrac{b}{\sqrt{b^2 - a^2}}

  2. ba\dfrac{b}{a}

  3. b2−a2b\dfrac{\sqrt{b^2 - a^2}}{b}

  4. ab2−a2\dfrac{a}{\sqrt{b^2 - a^2}}

Trigonometrical Ratios

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Answer

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

If sin θ = a/b, then cos θ is equal to? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

sin θ = PerpendicularHypotenuse\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}

Substituting values we get :

⇒ab=ABAC\Rightarrow \dfrac{a}{b} = \dfrac{AB}{AC}

Let AB = ak and AC = bk.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ (bk)2 = (ak)2 + BC2

⇒ b2k2 = a2k2 + BC2

⇒ BC2 = b2k2 - a2k2

⇒ BC2 = k2(b2−a2)\sqrt{k^2(b^2 - a^2)}

⇒ BC = k(b2−a2)k\sqrt{(b^2 - a^2)}.

By formula,

cos θ=BaseHypotenuse=BCAC=k(b2−a2)bk=b2−a2b\text{cos θ} = \dfrac{\text{Base}}{\text{Hypotenuse}} \\[1em] = \dfrac{BC}{AC} \\[1em] = \dfrac{k\sqrt{(b^2 - a^2)}}{bk} \\[1em] = \dfrac{\sqrt{b^2 - a^2}}{b}

Hence, Option 3 is the correct option.

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