Question

Without using trigonometric tables, prove that:

cosec 60° – sec 30° = 0

Answer 1

Given: cosec 60° – sec 30°

To Prove: cosec 60° – sec 30° = 0

Proof [Step-wise]:

1. Construct an equilateral triangle ABC with side length 2. (Each angle = 60°.)

2. Let AD be the altitude from A to BC.

D is the midpoint of BC.

So, BD = DC = 1 and AD ⟂ BC.

3. In right triangle ABD, AB = 2 and BD = 1.

By Pythagoras, `AD = sqrt(AB^2 - BD^2)` 

= `sqrt(4 - 1)`

= `sqrt(3)`

4. Angle B of triangle ABD is 60° and angle A in ABD is 30° (since triangle ABC is equilateral).

5. For angle 60° at B in triangle ABD, 

`sin 60^circ = "Opposite"/"Hypotenuse"`

= `(AD)/(AB)` 

= `(sqrt(3))/2`
For angle 30° at A in triangle ABD,

`cos 30^circ = "Adjacent"/"Hypotenuse"`

= `(AD)/(AB)`

= `(sqrt(3))/2`

Hence, sin 60° = cos 30°

= `(sqrt(3))/2`

6. Therefore, cosec 60° = `1/(sin 60^circ)` 

= `1/(sqrt(3)/2)`

= `2/sqrt(3)`

And `sec 30^circ = 1/(cos 30^circ)` 

= `1/(sqrt(3)/2)`

= `2/sqrt(3)`

7. Subtracting gives cosec 60° – sec 30°

= `2/sqrt(3) - 2/sqrt(3)`

= 0