If x=asecθcosϕ, y=bsecθcosϕand z=ctanθ, show that x²/a²+y²/b²-z²/c²=1
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Step-by-step explanation:
x = a secθ*cosΦ
squaring both sides we get,
→ x² = a² sec²θ*cos²Φ
Dividing by a² both sides,
→ x²/a² = sec²θ*cos²Φ --------Equation(1)
similarly,
y = bsecθ*sinΦ
→ y² = b²sec²θ*sin²Φ
→ y²/b² = sec²θ sin²Φ ---------Equation(2)
and,
z = ctanθ
→ z² = c²tan²θ
→ z²/c² = tan²θ ----------Equation(3)
Now, in LHS we have ,
Putting all values from Equation(1) , Equation(2) and Equation(3) we get,
→ x²/a² + y²/b² - z²/c²
→ sec²θ.cos²Φ + sec²θ sin²Φ - tan²θ
→ sec²θ(cos²Φ + sin²Φ) - tan²θ
Now, we know that sin²A + cos²A = 1
→ sec²θ*1 - tan²θ
→ sec²θ - tan²θ
Also, sec²A - tan²A = 1 .
→ 1 = RHS
✪✪ Hence Proved ✪✪
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