Is This Trigonometric Identity Valid for All Values?

[anemone]

Let $\dfrac{\cos^4 a}{x}+\dfrac{\sin^4 a}{y}=\dfrac{1}{x+y}$ for all real $a,\,b,\,x,\,y$.

Prove that $\dfrac{\cos^8 a}{x^3}+\dfrac{\sin^8 a}{y^3}=\dfrac{1}{(x+y)^3}$

 

[kaliprasad]

Let $\dfrac{\cos^4 a}{x}+\dfrac{\sin^4 a}{y}=\dfrac{1}{x+y}$ for all real $a,\,b,\,x,\,y$.

Prove that $\dfrac{\cos^8 a}{x^3}+\dfrac{\sin^8 a}{y^3}=\dfrac{1}{(x+y)^3}$

Spoiler

$\dfrac{\cos^4 a }{x} + \dfrac{\sin ^4 a }{y}= \dfrac{1}{x+y}$
hence
$\dfrac{\cos^4 a }{x} + \dfrac{(1-\cos^2 a )^2}{y}= \dfrac{1}{x+y}$
or
$\dfrac{\cos^4 a }{x} + \dfrac{1-2\cos^2 a +cos^4 a}{y}= \dfrac{1}{x+y}$
or
$(x+y)^2\cos^4a-2 x(x +y) \cos^2 a + x(x+y)= xy$
or $(x+y)^2\cos^4a -2 x(x +y) \cos^2 a + x^2= 0$
or $((x+y)\cos^2a -x)^2=0$
hence $\cos^2 a = \dfrac{x}{x+y}\cdots(1)$
from (1)
$\sin ^2 a = 1-\dfrac{x}{x+y}=\dfrac{y}{x+y}\cdots(2)$
using (1) and (2)
$\dfrac{\cos^8 a}{x^3} + \dfrac{\sin ^8 a}{y^3}$
= $\dfrac{(\cos^2 a)^4}{x^3} + \dfrac{(\sin ^2 a)^4}{y^3}$
= $\dfrac{(\frac{x}{x+y})^4}{x^3} + \dfrac{(\frac{y}{x+y})^4}{y^3}$
= $\dfrac{x}{(x+y)^4} + \dfrac{y}{(x+y)^4}$
= $\dfrac{x+y}{(x+y)^4}$
= $\dfrac{1}{(x+y)^3}$

 

[anemone]

Good job, kaliprasad!