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![[Post New]](/templates/default/images/icon_minipost_new.gif) 20 Jun 2008 15:13:12 IST
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1/(cosecA-cotA) -1/sinA=1/sinA-1/(cosecA+cotA)
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20 Jun 2008 15:48:33 IST
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posting the solution........
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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20 Jun 2008 15:55:19 IST
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LHS=
RHS=
As LHS=RHS so it is proved.
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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20 Jun 2008 23:25:42 IST
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This question can be solved more easily by rearranging......
1/ (cosecA - cot A) - 1 / sinA = 1/ sin A - 1/ (cosec A + cot A)
==> 1/ (cosec A - cot A) + 1/ (cosec A + cot A) = 2/ sin A ----1
now taking LHS
{ [cosec A + cot A] + [cosec A - cot A] } / (cosec^2A - cot^2A)
==> 2cosec A / 1
==> 2 / sinA =RHS of eq. 1
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20 Jun 2008 23:27:36 IST
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Great. Alternate ways are always welcomed.
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God does not care about our mathematical difficulties. He integrates empirically. ~~~Albert Einstein (1879-1955)~~~~
To divide a cube into two other cubes, a fourth power or in general any power whatever into two powers of the same denomination above the second is impossible, and I have assuredly found an admirable proof of this, but the margin is too narrow to contain it.~~~Pierre de Fermat (1601-1665)~~~
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