MCQ
If $sin\ x + cos\ x = a$, $a \in \left[ { - \sqrt 2 ,\sqrt 2 } \right] - \left\{ { - 1,1} \right\}$, then
$\sum\limits_{n = 1}^\infty  {\left( {{{\sin }^n}\ x + {{\cos }^n}\ x} \right)}$ is equal to -
  • A
    $\frac{{2(1 + a - {a^2})}}{{{{(a + 1)}^2}}}$
  • B
    $\frac{{2({a^2} - a + 1)}}{{{{(a - 1)}^2}}}$
  • C
    $\frac{{2({a^2} - a + 1)}}{{{{(a + 1)}^2}}}$
  • $\frac{{2(1 + a - {a^2})}}{{{{(a - 1)}^2}}}$

Answer

Correct option: D.
$\frac{{2(1 + a - {a^2})}}{{{{(a - 1)}^2}}}$
d
$\sum\limits_{n = 1}^\infty  {{{\sin }^n}x + } \sum\limits_{n = 1}^\infty  {{{\cos }^n}x}  = \frac{{\sin x}}{{1 - \sin x}} + \frac{{\cos x}}{{1 - \cos x}}$

$=\frac{[(\sin x+\cos x)-\sin 2 x]}{1-(\sin x+\cos x)+\sin x \cos x}$

Put $\sin 2 x=a^{2}-1 \,\,\, and \,\,\,  \sin x+\cos x=a$

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