If (x/a)cos theta + (y/b) sin theta = 1
And (x/a)sin theta - (y/b)cos theta = 1
Prove that (x²/a²) + (y²/b²)=2
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Answered by
16
Heya friends...✍
Here is ur answer...
☣☣☣☣☣☣☣☣☣☣☣☣☣☣☣☣
(x/a)cos¢+(y/b)sin¢=1------1)
(x/a)sin¢-(y/b)cos¢=1-------2)
first of all adding and squaring both equation..1)and 2)
we get. ....
【(x/a)cos¢+(y/b)sin¢)^2】+【(x/a)sin¢-y/bcos¢)^2】=1^2+1^2
=>x^2/a^2cos^2¢+y^2/b^2sin^2¢+2x/asin¢*y/b sin¢+x^2/a^2sin^2 ¢+y^2/a^2cos¢-x/a sin¢*y/bcos¢=1^2+1^2
=>x^2/a^2sin¢+y^2/b^2cos¢+x^2/a^2sin^2¢+y^2/b^2cos^2¢=2
=>x^2/a^2sin^2¢+x^2/a^2cos^2¢+y^2/b^2 sin^2¢+y^2/b^2cos^2¢=2
=>x^2/a^2(sin^2¢+cos^2¢)+y^2/b^2(cos^2¢+sin^2¢).=2
=>x^2/a^2+y^2/b^2=2. ....prooved here ..
hope it help you ..
@Rajukumar☺☺
Here is ur answer...
☣☣☣☣☣☣☣☣☣☣☣☣☣☣☣☣
(x/a)cos¢+(y/b)sin¢=1------1)
(x/a)sin¢-(y/b)cos¢=1-------2)
first of all adding and squaring both equation..1)and 2)
we get. ....
【(x/a)cos¢+(y/b)sin¢)^2】+【(x/a)sin¢-y/bcos¢)^2】=1^2+1^2
=>x^2/a^2cos^2¢+y^2/b^2sin^2¢+2x/asin¢*y/b sin¢+x^2/a^2sin^2 ¢+y^2/a^2cos¢-x/a sin¢*y/bcos¢=1^2+1^2
=>x^2/a^2sin¢+y^2/b^2cos¢+x^2/a^2sin^2¢+y^2/b^2cos^2¢=2
=>x^2/a^2sin^2¢+x^2/a^2cos^2¢+y^2/b^2 sin^2¢+y^2/b^2cos^2¢=2
=>x^2/a^2(sin^2¢+cos^2¢)+y^2/b^2(cos^2¢+sin^2¢).=2
=>x^2/a^2+y^2/b^2=2. ....prooved here ..
hope it help you ..
@Rajukumar☺☺
Answered by
0
According To Question ;
- and
- We need to Prove that
- Consider as Equation 1
- Consider as Equation 2
Now Square Equation 1 first and Equation 2 next on Both Sides
Add Equation 3 and Equation 4
Apply identity for expanding Equation 3, Where a = and b = and Apply identity for expanding Equation 4, Where a = and b =
When we expand Equation 3 and Equation 4 by identities, We get and
We know from Trigonometric Identities
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