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If tan A = xy\dfrac{x}{y}, then cos A is equal to

  1. xx2+y2\dfrac{x}{\sqrt{x^2 + y^2}}

  2. yx2+y2\dfrac{y}{\sqrt{x^2 + y^2}}

  3. x2−y2x2+y2\dfrac{x^2 - y^2}{\sqrt{x^2 + y^2}}

  4. x2−y2x2+y2\dfrac{x^2 - y^2}{x^2 + y^2}

Trigonometrical Ratios

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Answer

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

If tan A = x/y, then cos A is equal to? Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By formula,

tan A = PerpendicularBase\dfrac{\text{Perpendicular}}{\text{Base}}

Substituting values we get :

⇒xy=BCAB\Rightarrow \dfrac{x}{y} = \dfrac{BC}{AB}

Let BC = xk and AB = yk.

In right angle triangle ABC,

⇒ AC2 = AB2 + BC2

⇒ AC2 = (yk)2 + (xk)2

⇒ AC2 = y2k2 + x2k2

⇒ AC2 = k2(y2 + x2)

⇒ AC = k2(y2+x2)\sqrt{k^2(y^2 + x^2)}

⇒ AC = ky2+x2k\sqrt{y^2 + x^2}.

By formula,

cos A = BaseHypotenuse=ABAC=ykkx2+y2=yx2+y2\dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{AB}{AC} = \dfrac{yk}{k\sqrt{x^2 + y^2}} = \dfrac{y}{\sqrt{x^2 + y^2}}.

Hence, Option 2 is the correct option.

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