If x = asecθcos∅, y = bsecθsin∅ and z = ctanθ, then the value of (x²over a²+y²over b²-z²over c²) is equal to:

If x = asecθcos∅, y = bsecθsin∅ and z = ctanθ, then the…

If x = asecθcos∅, y = bsecθsin∅ and z = ctanθ, then the value of $({x^2\over a^2}+{y^2\over b^2}-{z^2\over c^2})$ is equal to:

Show answer & explanation

Correct answer: B

Explanation

The correct answer is option 2 i.e. 1.

x = a sec θ cos ∅

⇒ x/a = sec θ cos ∅ .....(1)

y = b sec θ sin ∅

⇒ y/b = sec θ sin ∅ .....(2)

z = c tan θ

⇒ z/c = tan θ ....(3)

Now, (x²/a²) + (y²/b²) - (z²/c²) = (x/a)² + (y/b)² - (z/c)²

= sec² θ cos² ∅ + sec² θ sin² ∅ - tan² θ

= sec² θ(cos² ∅ + sin² ∅) - tan² θ

= sec² θ - tan² θ [∵ cos²x + sin²x = 1]

= 1 [∵ sec²x - tan²x = 1]

Therefore, the value of the given expression is 1.