If x = a sec θ cos φ, y = b sec θ sin φ and z = c tan θ, then x^2/a^2 + y^2/b^2 =

Question

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 1

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]

= \(1+\frac{z^2}{c^2}\)