Question

Please solve with full explanation. ​

Answer 1

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QUESTION:-

$the \: equation \: \frac{ {x}^{2}}{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1 \: is \: true \: if \:$

(1) x = a tanθ and y = b cotθ

(2) x = a secθ and y = b cosecθ

(3) x = a cosθ and y = b sinθ

(4) x = a and y = b

SOLUTION:-

●firstly, we try to solve it with the help of 3 option...

→ $\frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }$

● Here x = a cosθ and y = b sinθ

→ $\frac{ { (a \: cosθ)}^{2} }{ {a}^{2} } + \frac{ {(b \: sinθ)}^{2} }{ {b}^{2} }$

→ $\frac{ {a}^{2} \: { \cosθ }^{2} }{ {a}^{2} } + \frac{{b}^{2}\: \ { \sinθ}^{2} }{ {b}^{2} }$

● so a^2 cancel out a^2 and b^2 cancel out b^2

→ ${ \cosθ }^{2} + { \sinθ }^{2}$

● and we know the identity cosθ^2 + sinθ^2 = 1

→ 1 = 1 ( L.H.S = R.H.S)

HENCE PROVED...

ANSWER:-

(3) x = a cosθ and y = b sinθ is correct option

Answer 2

Step-by-step explanation:

theequation

a

2

x

2

+

b

2

y

2

=1istrueif

(1) x = a tanθ and y = b cotθ

(2) x = a secθ and y = b cosecθ

(3) x = a cosθ and y = b sinθ

(4) x = a and y = b

SOLUTION:-

●firstly, we try to solve it with the help of 3 option...

→ \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} }

a

2

x

2

+

b

2

y

2

● Here x = a cosθ and y = b sinθ

→ \frac{ { (a \: cosθ)}^{2} }{ {a}^{2} } + \frac{ {(b \: sinθ)}^{2} }{ {b}^{2} }

a

2

(acosθ)

2

+

b

2

(bsinθ)

2

→ \frac{ {a}^{2} \: { \cosθ }^{2} }{ {a}^{2} } + \frac{{b}^{2}\: \ { \sinθ}^{2} }{ {b}^{2} }

a

2

a

2

cosθ

2

+

b

2

b

2

sinθ

2

● so a^2 cancel out a^2 and b^2 cancel out b^2

→ { \cosθ }^{2} + { \sinθ }^{2} cosθ

2

+sinθ

2

● and we know the identity cosθ^2 + sinθ^2 = 1

→ 1 = 1 ( L.H.S = R.H.S)

HENCE PROVED...

ANSWER:-

(3) x = a cosθ and y = b sinθ is correct option

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