I think mathematical induction is a good hint, and you should just work through it.

Alternatively, for a more straightforward solution, show that the initial condition implies that \( \sin^2 \theta = \frac { x} { x+y}, \cos^2 \theta = \frac { y } { x+y} \). Then, the subsequent condition follows directly.

@Gabriel Barros
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how to post a picture, eh?? its quite troublesome.... i have the solutn ready bt i just can't expose it due to the lack of attachment option...:P

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## Comments

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TopNewestI think mathematical induction is a good hint, and you should just work through it.

Alternatively, for a more straightforward solution, show that the initial condition implies that \( \sin^2 \theta = \frac { x} { x+y}, \cos^2 \theta = \frac { y } { x+y} \). Then, the subsequent condition follows directly.

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To prove the second condition is much neater.

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Wat is trigonometry

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It can be solved by MATHEMATICAL INDUCTION!!!!! Please see this technique uploaded by Sir Calvin in the blog!!!

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At first we take the given expression :- (Sin^{4}ө) /x + (cos^{4} ө)/y = 1/x+y Now write cosө in terms of sinө (Sin^{4} ө) /x + ( 1-sin^{2} ө)^{2} /y =1/x+y Take LCM => [ ysin^{4}ө + x(1-2sin^{2}ө + sin^{4}ө) ] /xy = 1/x+y Now cross multiply => sin^{4}ө (x+y)^{2} + x^{2} + xy – 2x^{2}sinө -2xysin^{2}ө = xy Let sin^ {2}ө = α => α^{2} (x+y)^{2} -2αx^{2} - 2xyα + x^{2} = 0 => α^{2}(x+y)^{2} -2αx ( x+y) +x^{2} = 0 =>[ α(x+y) - x ]^{2} = 0 => α = x / x+y We need to prove that sin^{2(m+1) }ө / x^{m} + cos^{2(m+1) }ө / y^{m} = 1/(x+y)^{m} At first we take the expression given on the left hand side :- => (sin^{2}ө )^{m+1} / x^{m} + (1-sin^{2}ө)^{m+1} / y^{m} But sin^{ 2}ө =α , as we had already assumed in the earlier steps =>( x/x+y)^{m+1} /x^{m} + (1 – x/x+y)^{m+1} /y^{m} => x^{m+1} / (x+y)^{m+1}x^{m} + y^{m+1} / (x+y)^{m+1} y^{m}

=> 1/(x+y)^{m+1} [ x^{m+1} / x^{m} + y^{m+1} / y^{m}] => 1/(x+y)^{m+1} [ x^{m+1} y^{m} + y^{m+1}x^{m} / (xy)^{m}] => 1/(x+y)^{m+1} [ x^{m}y^{m} (x+y) / x^{m}y^{m}] => (x+y) /(x+y)^{m+1} => ( x+y)^{–m} => 1/(x+y)^{m}

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its horrible to read that :P

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i have no idea regarding the latex form...so i cudn't convert to latex form bt u can go through this...and ask me fr ny help regarding math

contact me at :- prerana.chakrabarti@gmail.com

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is it done by mathematical induction....

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but how? :P

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arey....i was asking...from you....if it is done...by mathematical induction....i'll try...to xplain u....

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If m = 1 Then result verified

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we don't have to find the value of m...m is any integer for which the equation stands true..

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