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Powers with Negative Exponents

[ If e sin 2...

Question

$\begin{matrix} I f e^{({sin}^{2} x + {sin}^{2} x + {sin}^{6} x . . . . . . \infty) log e^{2}} a n d y^{2} - 5 y + 4 = 0 s a t i s f y t h e e q u a t i o n t h e n f i n d t h e v a l u e o f \frac{sin x}{cos x - sin x} \end{matrix}$

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Solution

$\begin{matrix} G i v e n e^{({sin}^{2} x + {sin}^{2} x + {sin}^{6} x . . . . . . \infty) log e^{2}} = e^{log 2 (\frac{1}{1 - {sin}^{2} x})} = 2^{(\frac{1}{1 - 2 {sin}^{2} x})} = 2^{{sec}^{2} x} = y N o w, y^{2} - 5 y + 4 = 0 \Rightarrow y^{2} - 4 y - y + 4 = 0 \Rightarrow y (y - 4) - 1 (y - 4) = 0 \Rightarrow (y - 4) (y - 1) = 0 ∴ y = 4 a n d y = 1 I f 2^{{sec}^{2} x} = 1 {sec}^{2} x = 0 (n o t p o s s i b l e a s s e c x \geq 1) e l s e, 2^{{sec}^{2} x} = 4 {sec}^{2} x = 2 cos x = \frac{1}{\sqrt{2}} T h u s, t h e v a l u e o f \frac{sin x}{cos x - sin x} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}} \Rightarrow N o t d e f i n e d . \end{matrix}$

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