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.)
Your second block of text...