If cos x + sec x = -2,then for a positive int | vedclass

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(C) Given $\cos x + \sec x = -2$.Since $\sec x = \frac{1}{\cos x}$,we have $\cos x + \frac{1}{\cos x} = -2$.Multiplying by $\cos x$,we get $\cos^2 x +

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$-2$ if $n$ is odd and $2$ if $n$ is even

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(C) Given $\cos x + \sec x = -2$.Since $\sec x = \frac{1}{\cos x}$,we have $\cos x + \frac{1}{\cos x} = -2$.Multiplying by $\cos x$,we get $\cos^2 x + 1 = -2 \cos x$.This simplifies to $\cos^2 x + 2 \cos x + 1 = 0$,which is $(\cos x + 1)^2 = 0$.Thus,$\cos x = -1$.Then $\sec x = \frac{1}{-1} = -1$.Now,$\cos^n x + \sec^n x = (-1)^n + (-1)^n = 2(-1)^n$.If $n$ is odd,$2(-1)^n = -2$.If $n$ is even,$2(-1)^n = 2$.Therefore,the value is $-2$ if $n$ is odd and $2$ if $n$ is even.

If $\cos x + \sec x = -2$,then for a positive integer $n$,$\cos^n x + \sec^n x$ is

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