Question

From the following figure, find:
(i) y
(ii) sin x°
(iii) (sec x° - tan x°) (sec x° + tan x°)

Answer 1

Consider the given figure :

(i) Since the triangle is a right-angled triangle, so using Pythagorean Theorem
2² = y² + 1²
y² = 4 – 1 = 3
y = ${\displaystyle \sqrt{3}}$ 

(ii) sin x° = ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}= \frac{\sqrt{3}}{2}}$

(iii) tan x° = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\sqrt{3}}$

     sec x° = ${\displaystyle \frac{\text{hypotenuse}}{\text{base}}=2}$

Therefore
( sec x° – tan x°) ( sec x° + tan x°)

= (2–${\displaystyle \sqrt{3}}$) (2+${\displaystyle \sqrt{3}}$)

= 4 – 3

= 1