Question

If sin α + cosec α = 2, find the value of sinⁿα + cosecⁿα, n ∈ Z.

Answer 1

given, $sin\alpha+cosec\alpha=2$

we know, $cosecant=\frac{1}{sine}$

then,
$sin\alpha+\frac{1}{sin\alpha}=2\\\\sin^2\alpha+1=2sin\alpha\\\\sin^2\alpha-2sin\alpha+1=0\\\\sin\alpha=1$

$\because sin\alpha=1\\\\\therefore cosec\alpha=1$

now, $sin^n\alpha+cosec^n\alpha=1^n+1^n=2$

hence, sinⁿα + cosecⁿα = 2

Answer 2

HELLO DEAR,



GIVEN:- sinα + cosecα = 2

=> sinα + 1/sinα = 2

=> sin²α + 1 = 2sinα

=> sin²α - 2sinα + 1 = 0

=> (sinα - 1)² = 0

=> sinα = 1

therefore, cosecα = 1


thus,
sinⁿα + cosecⁿα = 1ⁿ + 1ⁿ = 2 [n ∈ Z]


I HOPE IT'S HELP YOU DEAR,
THANKS