KnowledgeBoat Logo

Mathematics

If in ∆ABC, ∠C = 90° and tan A = 34\dfrac{3}{4}, prove that

sin A cos B + cos A sin B = 1.

Trigonometrical Ratios

28 Likes

Answer

Let ABC be a right angled triangle with ∠C = 90°.

If in ∆ABC, ∠C = 90° and tan A = 3/4, prove that sin A cos B + cos A sin B = 1. Trigonometrical Ratios, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Given,

tan A = 34\dfrac{3}{4}

By formula,

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

34=BCAC\dfrac{3}{4} = \dfrac{BC}{AC}

Let BC = 3x and AC = 4x.

In △ABC,

⇒ AB2 = AC2 + BC2

⇒ AB2 = (4x)2 + (3x)2

⇒ AB2 = 16x2 + 9x2

⇒ AB2 = 25x2

⇒ AB = 25x2\sqrt{25x^2}

⇒ AB = 5x.

By formula,

sin A=PerpendicularHypotenuse=BCAB=3x5x=35sin B=PerpendicularHypotenuse=ACAB=4x5x=45cos A=BaseHypotenuse=ACAB=4x5x=45cos B=BaseHypotenuse=BCAB=3x5x=35.\text{sin A} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{BC}{AB} = \dfrac{3x}{5x} = \dfrac{3}{5} \\[1em] \text{sin B} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{AC}{AB} = \dfrac{4x}{5x} = \dfrac{4}{5} \\[1em] \text{cos A} = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{AC}{AB} = \dfrac{4x}{5x} = \dfrac{4}{5} \\[1em] \text{cos B} = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{BC}{AB} = \dfrac{3x}{5x} = \dfrac{3}{5}.

Substituting values in L.H.S. of sin A cos B + cos A sin B = 1.

sin A cos B + cos A sin B=35×35+45×45=925+1625=9+1625=2525=1\text{sin A cos B + cos A sin B} = \dfrac{3}{5} \times \dfrac{3}{5} + \dfrac{4}{5} \times \dfrac{4}{5} \\[1em] = \dfrac{9}{25} + \dfrac{16}{25} \\[1em] = \dfrac{9 + 16}{25} \\[1em] = \dfrac{25}{25} \\[1em] = 1

Since, L.H.S. = R.H.S.

Hence, proved that sin A cos B + cos A sin B = 1.

Answered By

19 Likes


Related Questions