Question

In tan θ = 1, find the value of 5cot²θ + sin²θ - 1.

Answer 1

Consider ΔABC, where ∠A = 90°

tan θ = `"Perpendicular"/"Basse" = "AB"/"AC" = 1 = (1)/(1)`

By Pythagoras theorem,
BC²
= AB² + AC²
= 1² + 1²
= 2
⇒ BC = `sqrt(2)`
Now,

cot θ = `(1)/"tan θ"` = 1

sin θ = `"AB"/"BC" = (1)/sqrt(2)`

∴ 5cot²θ + sin²θ - 1

= `5 xx (1)^2 + (1/sqrt(2))^2 - 1`

= `5 + (1)/(2) - 1`

= `4 + (1)/(2)`

= `(9)/(2)`.