Question #5adf9

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Question #5adf9

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Answer 1 · 2016-04-03T14:59:20

Multiply by the conjugate.

Explanation:

$\frac{sin x - cos x}{sin x + cos x} \times \frac{sin x - cos x}{sin x - cos x}$ $(sinx - cosx) / (sinx + cosx) xx (sinx - cosx)/(sinx - cosx)$

$\frac{{sin}^{2} x - 2 cos x sin x + {cos}^{2} x}{{sin}^{2} x - {cos}^{2} x}$ $(sin^2x - 2cosxsinx + cos^2x)/(sin^2x - cos^2x)$

Now, you must apply the pythagorean identity ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2x = 1$

$\frac{1 - 2 cos x sin x}{{sin}^{2} x + {sin}^{2} x - 1}$ $(1 - 2cosxsinx)/(sin^2x + sin^2x - 1)$

$\frac{1 - 2 cos x sin x}{2 {sin}^{2} x - 1}$ $(1 - 2cosxsinx)/(2sin^2x - 1)$

Thus, you have your answer.

Hopefully this helps.

Answer 2 · 2016-04-03T16:57:19

$- cot (\frac{π}{4} + x)$ $-cot(pi/4+x)$ .

Explanation:

Numerator is $- \sqrt{2} (cos x cos (\frac{π}{4}) - sin x sin (\frac{π}{4})) = - cos (\frac{π}{4} + x)$ $-sqrt2(cos x cos (pi/4)-sinx sin (pi/4))=-cos(pi/4+x)$ .
Denominator is $\sqrt{2} (sin x cos (\frac{π}{4}) + cos x sin (\frac{π}{4})) = sin (\frac{π}{4} + x)$ $sqrt2(sin x cos (pi/4)+cos x sin (pi/4))=sin(pi/4+x)$ .
So, the ratio is $- cot (\frac{π}{4} + x)$ $-cot (pi/4+x)$ ..

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Question #5adf9

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