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Question #3a14d

1 Answer

Apr 14, 2017

Given
$cosecx+sinx=2$

$=>1/sinx+sinx=2$

$=>1+sin^2x=2sinx$

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

$=>(1-sinx)^2=0$

$=>1-sinx=0$

$=>sinx=1$

and $cosecx=1$

Now

$cosec^nx+sin^nx$

$=1^n+1^n=2$

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