if sin4A/a + cos4A/b = 1/ a+ b , then prove that sin8A/a3 + cos8A/b3 = 1/( a+b)3
Answers
||✪✪ QUESTION ✪✪||
if sin⁴A/a + cos⁴A/b = 1/ a+ b , then prove that = (sin⁸A/a³ + cos⁸A/b³) = 1/( a+b)³ ?
|| ✰✰ ANSWER ✰✰ ||
→ sin⁴A/a + cos⁴A/b = 1/ (a+ b)
Now, using cos²A = (1 - sin²A) or, cos⁴A = (1 - sin²A)² = 1 + sin⁴A - 2sin²A in LHS, we get,
→ sin⁴A/a + (1 + sin⁴A - 2sin²A)/b = 1/(a + b)
→ sin⁴A/a + 1/b + sin⁴A/b - 2sin²A/b = 1/(a + b)
→ sin⁴A(1/a + 1/b) -2sin²A/b = 1/(a + b) - 1/b
→ sin⁴A(a + b)/ab - 2asin²A/ab = (b - a - b )/(a + b)b
Cross - multiplying and cancelling Denominator b From both sides ,
→ (a + b)²(sin²A)² - 2a(a + b) sin²A = -(a)²
→ [(a + b)sin²A]² -2.a.(a + b)sin²A + a² = 0
Comparing it with a² - 2ab + b² = (a - b)² , we get,
→ [(a + b)sin²A - a]² = 0
→ sin²A = a/(a+b) ----------------- Equation (1)
→ 1 - sin²A = 1 - a/(a + b) = b/(a + b)
→ cos²A = b/(a + b) -------------- Equation (2)
Doing 4 Time both of The Equations both sides now, we get,
→ (sin²A)⁴ = [a/(a+b)]⁴
→ sin⁸A = a⁴ / (a+b)⁴
And,
→ (cos²A)⁴ = [ b/(a + b) ]⁴
→ cos⁸A = b⁴/(a+b)⁴
So,
→ (sin⁸A/a³) = [a/(a + b)⁴]
→ (cos⁸A/b³) = [b/(a + b)⁴]
So ,
→ (sin⁸A/a³ + cos⁸A/b³) = [a/(a + b)⁴] + [b/(a + b)⁴]
→ (sin⁸A/a³ + cos⁸A/b³) = (a + b)/(a + b)⁴