How do you rationalize the numerator for $\sqrt{\frac{sin x}{cot x}}$ $sqrt(sinx/cotx)$ ?

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Answer 1 · 2016-11-01T00:17:58

Simplify within the square root first, using the identity $cot x = \frac{cos x}{sin x}$ $cotx = cosx/sinx$ .

$\Rightarrow \sqrt{\frac{sin x}{\frac{cos x}{sin x}}}$ $=>sqrt(sinx/(cosx/sinx))$

$\Rightarrow \sqrt{\frac{{sin}^{2} x}{cos x}}$ $=>sqrt(sin^2x/cosx$

$\Rightarrow \frac{\sqrt{{sin}^{2} x}}{\sqrt{cos x}} \times \frac{\sqrt{cos x}}{\sqrt{cos x}}$ $=> sqrt(sin^2x)/sqrt(cosx) xx sqrt(cosx)/sqrt(cosx)$

$\Rightarrow \frac{\sqrt{{sin}^{2} x cos x}}{cos x}$ $=>sqrt(sin^2xcosx)/cosx$

Hopefully this helps!

How do you rationalize the numerator for sqrt(sinx/cotx)√sinxcotxsqrt(sinx/cotx)?