If sin⁴Θ/b + cos⁴Θ/a = (a+b)/ab , then prove that sin⁸Θ/a + cos⁸Θ/b = 1/(a³+b³)...

Let. theta.= x .

The given question is as follows:-

(sin^4 x)/a+(cos^4 x)/b = 1/(a+b). , prove that. (sin^8 x)/a^3 + (cos^8 x)/b^3

= 1/(a+b)^3.

(sin^4 x)/a +(cos ^4 x)/b= 1/(a+b)

or. b.(sin^4 x) +a.(1-sin^2 x)^2= a.b/(a+b)

or. b.sin^4 x+ a+a.sin^4 x-2a.sin^2 x= a.b/(a+b)

or. (a+b).sin^4 x-2.a.sin^2x+a = a.b/(a+b)

or. (a+b)^2.sin^4 x-2a(a+b).sin^2 x+a^2+ab=ab

or. {(a+b).sin^2 x}^2–2×(a+b).sin^2 x. × a. +(a)^2. = 0

or. [(a+b).sin^2 x. - a. ]^2. = 0

or. sin^2 x. = a/(a+b)……………..(1)

or. [ sin^2 x]^4. = a^4/(a+b)^4.

or. (sin^8 x)/a^3. = a/(a+b)^4……………(2)

Now. cos^2 x= 1- sin^2x= 1- a/(a+b)

cos^2 x= b/(a+b)…………………….(3)

or. [cos^2 x]^4. = b^4/(a+b)^4

or. (cos^8 x)/b^3. = b /(a+b)^4……………(4)

Adding eqn.(2) and (4)

(sin^8 x)/a^3+(cos^8 x)/b^3= (a+b)/(a+b)^4. = 1/(a+b)^3. Proved.