If 3 cos A = 4 sin A: find the value of :

4 cos² A - 3 sin² A + 2

Given: 3 cos A = 4 sin A

⇒ $\dfrac{\text{sin A}}{\text{cos A}} = \dfrac{3}{4}$

⇒ tan A = $\dfrac{3}{4}$

⇒ $\dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{3}{4}$

Let the perpendicular be 3k and base be 4k.

Using Pythagoras theorem, we get

Hypotenuse² = Perpendicular² + Base²

= (3k)² + (4k)²

= 9k² + 16k²

= 25k²

⇒ Hypotenuse = $\sqrt{25k^2}$

= 5k

sin A = $\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$ = $\dfrac{3k}{5k}$ = $\dfrac{3}{5}$

cos A = $\dfrac{\text{Base}}{\text{Hypotenuse}}$ = $\dfrac{4k}{5k}$ = $\dfrac{4}{5}$

Now the value of 4 cos² A - 3 sin² A + 2

$= 4 \times \Big(\dfrac{4}{5}\Big)^2 - 3 \times \Big(\dfrac{3}{5}\Big)^2 + 2\\[1em] = 4 \times \Big(\dfrac{16}{25}\Big) - 3 \times \Big(\dfrac{9}{25}\Big) + 2\\[1em] = \dfrac{64}{25} - \dfrac{27}{25} + 2\\[1em] = \dfrac{37}{25} + 2\\[1em] = \dfrac{37 + 50}{25}\\[1em] = \dfrac{87}{25}\\[1em] = 3\dfrac{12}{25}$

Hence, the value of 4 cos² A - 3 sin² A + 2 = $3\dfrac{12}{25}$.