If $\text{tan x} = 1\dfrac{1}{3}$, find the value of :

4 sin²x - 3 cos²x + 2

Given:

$\text{tan x} = 1\dfrac{1}{3} = \dfrac{4}{3}$

$⇒ \text{tan x} = \dfrac{Perpendicular}{Base} = \dfrac{4}{3}\\[1em]$

∴ If length of BC = 4a unit, length of AB = 3a unit.

In Δ ABC,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ AC² = (4a)² + (3a)²

⇒ AC² = 16a² + 9a²

⇒ AC² = 25a²

⇒ AC = $\sqrt{25 \text{a}^2}$

⇒ AC = 5a

sin x = $\dfrac{Perpendicular}{Hypotenuse}$

$= \dfrac{BC}{CA} = \dfrac{4a}{5a} = \dfrac{4}{5}$

cos x = $\dfrac{Base}{Hypotenuse}$

$= \dfrac{BA}{CA} = \dfrac{3a}{5a} = \dfrac{3}{5}$

Now,

4 sin²x - 3 cos²x + 2

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

Hence, 4 sin²x - 3 cos²x + 2 = $3\dfrac{12}{25}.$