If sin θ = 2/√5, find the value of sec^{2}θ + cot^{2}θ?

Option 1 : 21/4

**Given:**

Sinθ = 2/√5

**Concept:**

Using the trigonometric function, calculate the values of sec θ and cot θ.

**Calculation:**

sin θ = (Perpendicular)/(Hypotenuse)

In ∆ABC;

(AC)^{2} = (AB)^{2} + (BC)^{2}

⇒ (√5)^{2} = (2)^{2} + (BC)^{2}

⇒ 5 = 4 + (BC)^{2}

⇒ (BC)^{2} = 5 – 4

⇒ (BC)^{2} = 1

⇒ BC = 1

sec θ = (Hypotenuse)/Base)

⇒ sec θ = √5/1 = √5

cot θ = (Base)/(perpendicular)

⇒ cot θ = 1/2

(sec θ)^{2} + (cot θ)^{2} = (√5)^{2} + (1/2)^{2}

⇒ 5 + 1/4

⇒ 21/4

**∴ The value of sec2θ + cot2θ is 21/4.**