If sin x + cos x = a,where a in [-sqrt{2}, sq | vedclass

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(D) Given $\sin x + \cos x = a$. Squaring both sides,we get $1 + 2 \sin x \cos x = a^2$,so $\sin x \cos x = \frac{a^2 - 1}{2}$.The sum is $S = \sum_{n

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$\frac{2(1 + a - a^2)}{(a - 1)^2}$

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(D) Given $\sin x + \cos x = a$. Squaring both sides,we get $1 + 2 \sin x \cos x = a^2$,so $\sin x \cos x = \frac{a^2 - 1}{2}$.The sum is $S = \sum_{n=1}^{\infty} \sin^n x + \sum_{n=1}^{\infty} \cos^n x = \frac{\sin x}{1 - \sin x} + \frac{\cos x}{1 - \cos x}$.$S = \frac{\sin x(1 - \cos x) + \cos x(1 - \sin x)}{(1 - \sin x)(1 - \cos x)} = \frac{(\sin x + \cos x) - 2 \sin x \cos x}{1 - (\sin x + \cos x) + \sin x \cos x}$.Substituting $\sin x + \cos x = a$ and $\sin x \cos x = \frac{a^2 - 1}{2}$:$S = \frac{a - (a^2 - 1)}{1 - a + \frac{a^2 - 1}{2}} = \frac{a - a^2 + 1}{\frac{2 - 2a + a^2 - 1}{2}} = \frac{2(1 + a - a^2)}{a^2 - 2a + 1} = \frac{2(1 + a - a^2)}{(a - 1)^2}$.

If $\sin x + \cos x = a$,where $a \in [-\sqrt{2}, \sqrt{2}] - \{-1, 1\}$,then $\sum_{n=1}^{\infty} (\sin^n x + \cos^n x)$ is equal to -

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