Hey help me with this trigo proving question.
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Answered by
2
I've got two equation given to me
x/a cos © - y/b sin © = 1
Let's square this on both sides
Yo !!
x^2/a^2 cos ^2 © + y^2/b^2 sin ^2 © - 2xy/ab sin© cos © = 1. ..........(1)
Now,
x/a sin © - y/b cos © = 1
Squaring this on both sides,
x^2/a^2 sin^2 © + y^2/b^2 cos^2 © + 2xy/ab sin© cos © = 1.........(2)
Adding equations 1 and 2
X^2/a^2 sin^2 © + y^2/b^2 cos^2 © + x^2/a^2 cos^2 © + y^2/b^2 sin^2 © + 2xy/ab sin©cos© - 2xy/ab sin© cos © = 1+1
So,
x^2/a^2 ( sin^2 © + cos^2 ©) + y^2/b^2 ( sin^2 © + cos ^2 ©) = 2
Therefore,
x^2/a^2 + y^2/b^2 = 2
Hope this helps you !
x/a cos © - y/b sin © = 1
Let's square this on both sides
Yo !!
x^2/a^2 cos ^2 © + y^2/b^2 sin ^2 © - 2xy/ab sin© cos © = 1. ..........(1)
Now,
x/a sin © - y/b cos © = 1
Squaring this on both sides,
x^2/a^2 sin^2 © + y^2/b^2 cos^2 © + 2xy/ab sin© cos © = 1.........(2)
Adding equations 1 and 2
X^2/a^2 sin^2 © + y^2/b^2 cos^2 © + x^2/a^2 cos^2 © + y^2/b^2 sin^2 © + 2xy/ab sin©cos© - 2xy/ab sin© cos © = 1+1
So,
x^2/a^2 ( sin^2 © + cos^2 ©) + y^2/b^2 ( sin^2 © + cos ^2 ©) = 2
Therefore,
x^2/a^2 + y^2/b^2 = 2
Hope this helps you !
Answered by
0
Answer:
I've got two equation given to me
x/a cos © - y/b sin © = 1
Let's square this on both sides
Yo !!
x^2/a^2 cos ^2 © + y^2/b^2 sin ^2 © - 2xy/ab sin© cos © = 1. ..........(1)
Now,
x/a sin © - y/b cos © = 1
Squaring this on both sides,
x^2/a^2 sin^2 © + y^2/b^2 cos^2 © + 2xy/ab sin© cos © = 1.........(2)
Adding equations 1 and 2
X^2/a^2 sin^2 © + y^2/b^2 cos^2 © + x^2/a^2 cos^2 © + y^2/b^2 sin^2 © + 2xy/ab sin©cos© - 2xy/ab sin© cos © = 1+1
So,
x^2/a^2 ( sin^2 © + cos^2 ©) + y^2/b^2 ( sin^2 © + cos ^2 ©) = 2
Therefore,
x^2/a^2 + y^2/b^2 = 2
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