If $sin x + "cosec" x = 2$ , then find $sin^n x + "cosec" ^n x$ ?

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Answer 1 · 2016-09-03T10:46:07

$sin^nx + 1/(sin^n x) = 2$

Solving

$sin x + 1/sin x = 2 -> sin^2 x - 2 sinx + 1 = 0$

we have

$sin x = 1$ then

$sin^nx + 1/(sin^n x) = 2$

This can be also demonstrated by finite induction.

1) First it is true for $n=1$ (with sin x = 1 of course)
2) Supposing that it is true for $n$ then
3) Prove that it is true for $n+1$

It is quite easy and it is left as an exercise.

If sin x + "cosec" x = 2sin x + "cosec" x = 2, then find sin^n x + "cosec" ^n xsin^n x + "cosec" ^n x ?