Question

prove that x^4-y^4=b^2-a^2, if a cot(theta)+b cosec(theta)= x^2 and b cot(theta) + a cose(theta)= y^2 ...??????

Answer 1

sm1 help

Answer 2

Tell me if you understand this or not. :)

Answer 3

\[x ^{4}=(acot \theta +b \csc \theta)^{2}\]
\[y ^{4}=(acsc \theta + b \cot \theta)^{2}\]
\[x ^{4}=a^{2}\cot^{2} \theta + abcot \theta \csc \theta + b^{2}\csc^{2} \theta \]
\[y^{4} = a^{2}\csc^{2} \theta + abcot \theta \csc \theta + b^{2}\cot \theta\]
x^4 - y^4 = \[a^{2}\cot^{2} \theta + abcot \theta \csc \theta + b^{2} \csc^{2} \theta - a^{2}\csc^{2} \theta - abcot \theta \csc \theta - b^{2}\cot^{2} \theta\]
becomes:
\[a^{2}\cot^{2} \theta - a^{2} \csc^{2} \theta+b^{2}\csc^{2} \theta - b^{2}\cot^{2} \theta\]
\[a^{2}(\cot^{2} \theta - \csc^{2} \theta ) + b^{2}(\csc^{2} \theta - \cot^{2} \theta)\]
Pythagorean identity says 1 + cot^2 = csc^2
Therefore cot^2(theta) - csc^2(theta) = -1
and csc^2(theta) - cot^2(theta) = 1
\[a^{2}(-1) + b^{2}(1) \implies b^{2} - a^{2}\]
Therefore
\[x^{4} -y^{4} = b^{2} - a^{2}\]

Answer 4

And I have no way to do work on paper and take a pic like akash, lol. Taking reliable pictures of anything is not something I can do x_x

Answer 5

:D.... thnx anyway @Psymon .......and thnx @AkashdeepDeb ....i understud

Answer 6

:)
He DESERVES the medal. :)

Answer 7

Lol, you don't need my gibberish then.

Answer 8

Why do I?

Answer 9

LaTeX is tougher than writing. :P

Answer 10

I don't have the ability to write, I have to use latex xD

Answer 11

i knw......

Answer 12

wish i could give bth of u a medal...... :D

Answer 13

He answered it just as well if not better and faster, I don't deserve any medal, lol.