Question:
if SIN X COSEC X = 2, then SIN^ n X COSEC ^n X =
Answer:
Given,
sin x + cosec x = 2
=> sin x + 1/sin x = 2
=> (sin2 x + 1)/sin x = 2
=> sin2 x + 1 = 2sin x
=> sin2 x - 2sin x + 1 = 0
=> (sin x - 1)2 = 0
=> sin x - 1 = 0
=> sin x = 1
Now, sinn x + cosecn x = sinn x + 1/sinn x
= 1n + 1/1n
= 1 + 1
= 2
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