Question

85 th qes, pls answer fast

Answer 1

hyy mate ✌️

hope it will help you ❣️

Answer 2

Question :--- if x = a secθ*cosΦ , y = bsecθ*sinΦ and z = ctanθ , show that, x²/a² + y²/b² - z²/c² = 1.

Solution :---

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 ✪✪