Question

Find sec x + tan x

Answer 1

Given:

Triangle △PQS

PS = 16 cm (hypotenuse)

QS = 12 cm (base)

PQ = ?

Step 1: Use Pythagoras Theorem in △PQS

`PS^2 = PQ^2 + QS^2`

`16^2 = PQ^2 + 12^2`

256 = PQ² + 144

PQ² = 256 − 144 = 112

PQ = `sqrt112`

`sqrt 112 = sqrt(16 xx7) = 4sqrt7`

PQ = `4 sqrt 7`

Step 2: Use formulas

`sec x = "hypotenuse"/"adjacent" = (PS)/(QS) = 16/12 = 4/3`

`tan x = "Opposite"/"adjacent" = (PQ)/(QS) = 4sqrt7/12 = sqrt7/3`

`sec x + tan x = 4/3 + sqrt7/3 = (4 + sqrt7)/3`