if sin^4theta/a+cos^2theta/b =1/a+b .. then prove that sin^8theta / a^3+cos^8theta /b^3=1/a+b
Answers
Step-by-step explanation:
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.
Answer:
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