If θ is an acute angle and sin θ = cos θ, find the value of

2 tan² θ + sin² θ - 1.

Given,

sin θ = cos θ

⇒ $\dfrac{\text{sin θ}}{\text{cos θ}} = 1$

⇒ tan θ = 1.

⇒ tan θ = tan 45°

⇒ θ = 45°.

⇒ sin 45° = cos 45° = $\dfrac{1}{\sqrt{2}}$

Substituting values in 2 tan² θ + sin² θ - 1 we get :

$\Rightarrow 2(1)^2 + \Big(\dfrac{1}{\sqrt{2}}\Big)^2 - 1 \\[1em] \Rightarrow 2 + \dfrac{1}{2} - 1 \\[1em] \Rightarrow 1 + \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{2 + 1}{2} \Rightarrow \dfrac{3}{2}.$

Hence, 2 tan² θ + sin² θ - 1 = $\dfrac{3}{2}$.