Answer:
HELLO DEAR,
GIVEN:-
10sin⁴A + 15cos⁴A = 6
=> 10(sin²A)² + 15cos⁴A = 6
=> 10{(1 - cos²A)²} + 15cos⁴A = 6
=> 10{1 + cos⁴A - 2cos²A} + 15cos⁴A = 6
=> 10 + 10cos⁴A - 20cos²A + 15cos⁴A = 6
=> 25cos⁴A - 20cos²A + 4 = 0
=> 25cos⁴A - 10cos²A - 10cos²A + 4 = 0
=> 5cos²A(5cos²A - 2) - 2(5cos²A - 2) = 0
=> (5cos²A - 2)(5cos²A - 2) = 0
=> cos²A = 2/5
=> cosA = √2/√5 [ secA = √5/√2]
therefore,
sinA = √[1 - 2/5] = √3/√5 [ cosecA = √5/√3]
now,
27cosec^6A + 8sec^6A
=> 27 × {(√5/√3)²}³ + 8 × {(√5/√2)²}³
=> 27 × 125/27 + 8 × 125/8
=> 125 + 125
=> 250.
HENCE, 27cosec^6A + 8sec^6A = 250.
I HOPE IT'S HELP YOU DEAR,
THANKS
Step-by-step explanation:
