If x = a sec θ + b tan θ and y = a tan θ + b sec θ , then x 2 − y 2 a 2 − b 2 = ___ Options: A . &nbsp2 B . &nbsp1 C . &nbsp0 D . &nbsp- 1

If x=a sec θ+b tan θ and y=a tan

Question

x = a sec θ + b tan θ

and

y = a tan θ + b sec θ

, then

\frac{x^{2} - y^{2}}{a^{2} - b^{2}} =

___

Options:

A . 2

B . 1

C . 0

D . - 1

Answer: Option B
:
B

x = a sec θ + b tan θ, y = a tan θ + b sec θ

Squaring both sides we get,

x^{2} = (a sec θ + b tan θ)^{2}, y^{2} = (a tan θ + b sec θ)^{2}

\begin{matrix} ∴ & x^{2} = a^{2} {sec}^{2} θ + 2 a b sec θ tan θ + b^{2} {tan}^{2} θ y^{2} = a^{2} {tan}^{2} θ + 2 a b sec θ tan θ + b^{2} {sec}^{2} θ - - - - ------------------------------------------- - Subtracting, & x^{2} - y^{2} = a^{2} ({sec}^{2} θ - {tan}^{2} θ) + b^{2} ({tan}^{2} θ - {sec}^{2} θ) \end{matrix}

x^{2} - y^{2} = a^{2} ({sec}^{2} θ - {tan}^{2} θ) - b^{2} {(sec}^{2} θ - {tan}^{2} θ)

x^{2} - y^{2} = a^{2} - b^{2} \dots \dots ({sec}^{2} θ - {tan}^{2} θ = 1)

∴ \frac{x^{2} - y^{2}}{a^{2} - b^{2}} = 1

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