]> Exercise 2.6

## Exercise 2.6

Derive the relations between these functions by using similar triangles.

Solution:

By similar triangles, the ratio of BD to BO, $sin ⁡ θ$ to 1 is the same as the ratio of CB to OC, $tan ⁡ θ$ to $sec ⁡ θ$ , and of OB to AO (1 to $csc ⁡ θ$ ), and of CO to AC ( $s e c θ$ to ( $tan ⁡ θ + cot ⁡ θ$ )), and of CD to BC (( $sec ⁡ θ − cos ⁡ θ$ ) to $tan ⁡ θ$ ), and of OE to BO ( $sin ⁡ θ$ to 1)

This tells us

$csc ⁡ θ = 1 sin ⁡ θ$

so that on taking complements we have as well,

$sec θ = 1 cos ⁡ θ$

We also get

$tan ⁡ θ = sin ⁡ θ sec ⁡ θ = sin ⁡ θ cos ⁡ θ$

and the complementary result

$cot ⁡ θ = cos ⁡ θ sin ⁡ θ$