Question

if 10sin^4A + 15cos^4A = 6 then 27cosec^6A + 8sec^6A = ?​

Answer 1

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

Answer 2

Answer:

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