If x = a sec θ cos φ, y = b sec θ sin φ and z = c tan θ, then =

1.	If x = a sec θ cos φ, y = b sec θ sin φ and z = c tan θ, then = \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) A. \(\frac{z^2}{c^2}\) B. \(1-\frac{z^2}{c^2}\) C. \(\frac{z^2}{c^2}-1\) D. \(1+\frac{z^2}{c^2}\)
Answer» Given: x a sec θ cos ϕ Squaring both sides, we get x² = a²sec²θ cos²ϕ and y = b sec θ sin ϕ Squaring both sides, we get y² = b² sec²θ sin²ϕ And z = c tan θ ⇒ z² = c²tan²θ ⇒ tan²θ = \(\frac{z^2}{c^2}\) ........(i) To find: \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) Consider \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = \(\frac{a^2sec^2θcos^2Φ}{a^2} + \frac{b^2sec^2θsin^2Φ}{b^2}\) = sec²θ cos²ϕ + sec²θ sin²ϕ = sec²θ (cos²ϕ + sin²ϕ) = sec²θ [∵ sin²ϕ + cos²ϕ = 1] = 1 + tan²θ [∵ 1 + tan²θ = sec²θ] = \(1+\frac{z^2}{c^2}\)

If x = a sec θ cos φ, y = b sec θ sin φ and z = c tan θ, then = \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) A. \(\frac{z^2}{c^2}\) B. \(1-\frac{z^2}{c^2}\) C. \(\frac{z^2}{c^2}-1\) D. \(1+\frac{z^2}{c^2}\)

Answer»

Given: x a sec θ cos ϕ

Squaring both sides, we get

x² = a²sec²θ cos²ϕ

and y = b sec θ sin ϕ

Squaring both sides, we get

y² = b² sec²θ sin²ϕ

And z = c tan θ

⇒ z² = c²tan²θ

⇒ tan²θ = \(\frac{z^2}{c^2}\) ........(i)

To find: \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\)

Consider \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = \(\frac{a^2sec^2θcos^2Φ}{a^2} + \frac{b^2sec^2θsin^2Φ}{b^2}\)

= sec²θ cos²ϕ + sec²θ sin²ϕ

= sec²θ (cos²ϕ + sin²ϕ)

= sec²θ [∵ sin²ϕ + cos²ϕ = 1]

= 1 + tan²θ [∵ 1 + tan²θ = sec²θ]

= \(1+\frac{z^2}{c^2}\)