If sin4x/a+cos4x/b = 1/a+b. Prove that …

Given : Sin⁴x/a + cos⁴x/b = 1/a+b

To prove :
sin⁸x/a³ + cos⁸x/b³ =?

Solution:

Sin⁴x/a + cos⁴x/b = 1/a+b

As we know that Sin²x +Cos²x=1
∴Cos²x=1-sin²x
Cos⁴x=(Cos²x)²=(1-sin²x)²=1+sin⁴x-2sin²x

⇒Sin⁴x/a  +[1+sin⁴x-2sin²x]/b =1/a+b

⇒Sin⁴x/a+1/b + sin⁴x/b -2sin²x/b=(1/a+b)

⇒Sin⁴x/a+ sin⁴x/b -2sin²x/b=(1/a+b)-1/b

⇒Sin⁴x(1/a +1/b)-2sin²xa/ab=[b-a-b]/b(a+b)

⇒Sin⁴x(1/a +1/b)-2asin²x/ab=-a/b(a+b)

⇒Sin⁴x(a+b)/ab-2asin²x/ab=-a/b(a+b)


⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x= - a²

⇒(a+b)²(Sin²x)² - 2a(a+b)sin²x +  a² =0
This equation is similar to a²-2ab+b²=0 ⇒(a-b)²
[(a+b) Sin²x-a]²=0
sin²x=a/a+b

Cos²x=1-sin²x= 1- a/(a+b)= b/a+b
∴sin²x=a/a+b and Cos²x= b/a+b

Now :
sin⁸x=a⁴/(a+b)⁴  and Cos⁸x= b⁴/(a+b)⁴

sin⁸x/a³= a/(a+b)⁴   and Cos⁸x/b³=b/(a+b)⁴
Now 

sin⁸x/a³ + Cos⁸x/b³ =   a/(a+b)⁴ + b/(a+b)⁴
=a+b/(a+b)⁴
⇒sin⁸x/a³ + Cos⁸x/b³=1/(a+b)³