Question

Find sin θ + cos θ, and sec α + tan α in the following figure.

Answer 1

Given:

In triangle △ABD:

AD = 8 cm

BD = 6 cm

We are to find:

sin⁡θ + cos⁡θ 

In triangle △ADC:

AD = 8 cm

DC = 15 cm

We are to find:

sec α + tan α

Using Pythagoras Theorem:

`AB = sqrt (AD^2 + BD^2) = sqrt (8^2 + 6^2) = sqrt (64 + 36) = sqrt100 = 10cm`

`sin θ = (AD)/(AB) = 8/10 = 4/5`

`cos θ = (BD)/(AB) = 6/10 = 3/5`

∴ sin θ + cos θ = `4/5 + 3/5 = 7/5`

Using Pythagoras Theorem:

`AC = sqrt(AD^2 + DC^2) = sqrt(8^2 + 15^2) = sqrt(64 + 225) = sqrt289 = 17`

`sec α = (AC)/(AD) = 17/8`

`tan α = (DC)/(AD) = 15/8`

∴ sec α + tan α = `(17 + 15)/8 = 32/8 = 4`