If cot θ = 1; find the value of :

5 tan²θ + 2 sin²θ - 3

Given:

cot θ = 1

$⇒ \text{cot θ} = \dfrac{Base}{Perpendicular} = 1 \\[1em]$

∴ If length of AB = x unit, length of BC = x unit.

In Δ ABC,

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

⇒ AC² = (x)² + (x)²

⇒ AC² = x² + x²

⇒ AC² = 2x²

⇒ AC = $\sqrt{2\text{x}^2}$

⇒ AC = $\sqrt{2}$x

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

$= \dfrac{BC}{CA} = \dfrac{x}{\sqrt{2}x} = \dfrac{1}{\sqrt{2}}$

tan θ = $\dfrac{Perpendicular}{Base}$

$= \dfrac{BC}{BA} = \dfrac{x}{x} = 1$

Now,

5 tan² θ + 2 sin² θ - 3

$= 5 \times 1^2 + 2 \times \Big(\dfrac{1}{\sqrt{2}}\Big)^2 - 3\\[1em] = 5 + 2 \times \dfrac{1}{2} - 3\\[1em] = 5 + \cancel{2} \times \dfrac{1}{\cancel{2}} - 3\\[1em] = 5 + 1 - 3\\[1em] = 6 - 3\\[1em] = 3$

Hence, 5 tan²θ + 2 sin²θ - 3 = 3.