Question

if x = a sin thetay = b tan theta prove a²/x²-b²y²=1​

Answer 1

Step-by-step explanation:

ANSWER

We have ,

L.H.S=

x

2

a

2

−

y

2

b

2

⇒L.H.S=

a

2

sin

2

θ

a

2

−

b

2

tan

2

θ

b

2

[∵x=asinθ,y=btanθ]

⇒L.H.S=

sin

2

θ

1

−

tan

2

θ

1

⇒L.H.S=cosec

2

θ−cot

2

θ [∵1+cot

2

θ=cosec

2

θ∴cosec

2

θ−cot

2

θ=1]

⇒ LHS =1= RHS

Hence, proved

Answer 2

Answer:

$x = a \: sinθ \\ \frac{x}{a} = sinθ \\ squaring \: on \: both \: the \: sides \\ { (\frac{x}{a} )}^{2} = {(sinθ)}^{2} \\ \frac{ {x}^{2} }{ {a}^{2} } = {sin}^{2} θ \\ \frac{ {a}^{2} }{ {x}^{2} } = \frac{1}{ {sin}^{2}θ } \\ y = b \: tan θ \\ \frac{y}{b} = tanθ \\ squaring \: on \: both \: the \: sides \\ { (\frac{y}{b} )}^{2} = {(tanθ)}^{2} \\ \frac{ {y}^{2} }{ {b}^{2} } = {tan}^{2} θ \\ \frac{ {b}^{2} }{ {y}^{2} } = \frac{1}{ {tan}^{2}θ } \\ \frac{ {a}^{2} }{ {x}^{2} } - \frac{ {b}^{2} }{ {y}^{2} } = 1 \\ \frac{1}{ {sin}^{2}θ } - \frac{1}{ {tan}^{2}θ } = 1\\ \frac{1}{ {sin}^{2} θ} - \frac{1}{ \frac{ {sin}^{2}θ }{ {cos}^{2}θ } } = 1 \\ \frac{1}{ {sin}^{2}θ } - \frac{ {cos}^{2}θ }{ {sin}^{2} θ} = 1\\ \frac{1 - {cos}^{2} θ}{ {sin}^{2} θ} = 1\\ \frac{ {sin}^{2} θ}{ {sin}^{2} θ} = 1 \\LHS=RHS \\ Hence \: proved$