# CLASS-10TRIGONOMETRY - ELIMINATION OF TRIGONOMETRIC RATIOS

ELIMINATION OF TRIGONOMETRIC RATIOS -

Example.1)  If (cos θ + sin θ) = √2 cos θ, then prove that (cos θ – sin θ) = √2 sin θ

Ans.) We have, (cos θ + sin θ) = √2 cos θ

=>  (cos θ + sin θ)² = (√2 cos θ)²   [squaring both the side]

=>  cos² θ + sin² θ + 2. sin θ. cos θ =  2 cos² θ

=>  2 cos² θ – cos² θ - sin² θ = 2. sin θ. cos θ

=>  cos² θ - sin² θ = 2. sin θ. cos θ

=>  (cos θ + sin θ) (cos θ - sin θ) = 2. sin θ. cos θ

=>  √2 cos θ (cos θ – sin θ) = 2 sin θ . cos θ

[substitute (cos θ + sin θ) = √2 cos θ]

=>    (cos θ – sin θ) = √2 sin θ           (Proved)

Example.2) Eliminate θ between the following equations –

x = q tan θ + p sec θ , and y = p tan θ + q sec θ

Ans.)  The given equations are –

x = q tan θ + p sec θ ......................(i)

y = p tan θ + q sec θ ......................(ii)

on squaring (i) and (ii) and subtracting, we get –

x²- y² = (q tan θ + p sec θ)²- (p tan θ + q sec θ)

=  q² tan² θ + p² sec² θ + 2pq tan θ. sec θ – (p² tan² θ + q² sec² θ + 2pq tan θ. sec θ)

=  q² tan² θ + p² sec² θ + 2pq tan θ. sec θ – p² tan² θ - q² sec² θ - 2pq tan θ. sec θ

=   q² tan² θ + p² sec² θ – p² tan² θ - q² sec² θ

=   (q² - p²) tan² θ – (q² - p²) sec² θ

=   (q² - p²) (tan² θ - sec² θ)

=   (q² - p²) (- 1) (sec² θ - tan²θ)

=   (p² - q²) . 1      [As per formulae, sec² A - tan² A = 1]

=   (p² - q²)             (Ans.)