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

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

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

Given,

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

By formula,

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

⇒ $\dfrac{3}{4} = \dfrac{BC}{AC}$

Let BC = 3x and AC = 4x.

In △ABC,

⇒ AB² = AC² + BC²

⇒ AB² = (4x)² + (3x)²

⇒ AB² = 16x² + 9x²

⇒ AB² = 25x²

⇒ AB = $\sqrt{25x^2}$

⇒ AB = 5x.

By formula,

$\text{sin A} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{BC}{AB} = \dfrac{3x}{5x} = \dfrac{3}{5} \\[1em] \text{sin B} = \dfrac{\text{Perpendicular}}{\text{Hypotenuse}} = \dfrac{AC}{AB} = \dfrac{4x}{5x} = \dfrac{4}{5} \\[1em] \text{cos A} = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{AC}{AB} = \dfrac{4x}{5x} = \dfrac{4}{5} \\[1em] \text{cos B} = \dfrac{\text{Base}}{\text{Hypotenuse}} = \dfrac{BC}{AB} = \dfrac{3x}{5x} = \dfrac{3}{5}.$

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

$\text{sin A cos B + cos A sin B} = \dfrac{3}{5} \times \dfrac{3}{5} + \dfrac{4}{5} \times \dfrac{4}{5} \\[1em] = \dfrac{9}{25} + \dfrac{16}{25} \\[1em] = \dfrac{9 + 16}{25} \\[1em] = \dfrac{25}{25} \\[1em] = 1$

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

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