Question #f68d0

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Question #f68d0

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Answer 1 · 2017-03-23T23:11:13

$sec x - sin x tan x$ $secx-sinxtanx$

Explanation:

Since $csc x = \frac{1}{sin x}$ $cscx=1/sinx$ and $cot x = \frac{cos x}{sin x}$ $cotx=cosx/sinx$ , substitute these in to get $\frac{\frac{1}{sin x} - sin x}{\frac{cos x}{sin x}}$ $(1/sinx - sinx)/(cosx/sinx)$

Multiply the reciprocal of the denominator.

$\frac{1}{sin x} - sin x \cdot \frac{sin x}{cos x}$ $1/sinx - sinx*sinx/cosx$

We end up with this after using the distributive property.

$\frac{sin x}{sin x cos x} - \frac{{sin}^{2} x}{cos x}$ $sinx/(sinxcosx) - sin^2x/cosx$

Cancel out the two $sin x$ $sinx$ s on the left.

$\frac{1}{cos x} - \frac{{sin}^{2} x}{cos x}$ $1/cosx-sin^2x/cosx$

Since $\frac{1}{cos x} = sec x$ $1/cosx=secx$ and $\frac{sin x}{cos x} = tan x$ $sinx/cosx=tanx$ , the simplified answer is

$sec x - sin x tan x$ $secx-sinxtanx$

EDIT: Ignore this, not fully simplified. See Scott's answer.

Answer 2 · 2017-03-23T23:23:08

$cos x$ $cos x$

Explanation:

First put everything in terms of sine and cosine

$csc x = \frac{1}{sin x}$ $csc x = 1/sin x$ and $cot x = \frac{cos x}{sin x}$ $cot x = cos x/sin x$ , so $\frac{csc x - sin x}{cot x} = \frac{\frac{1}{sin x} - sin x}{\frac{cos x}{sin x}} = (\frac{1}{sin x} - sin x) (\frac{sin x}{cos x})$ $(csc x - sin x)/cot x = (1/sin x - sin x)/(cos x/sin x) = (1/sin x - sin x)(sin x/cos x)$

Then distribute multiplication over subtraction

$(\frac{1}{sin x} - sin x) (\frac{sin x}{cos x}) = (\frac{1}{sin x} \cdot \frac{sin x}{cos x}) - (sin x \cdot \frac{sin x}{cos x})$ $(1/sin x - sin x)(sin x/cos x) = (1/sin x * sin x/cos x) - (sin x * sin x/cos x)$

Multiply inside the parentheses

$(\frac{1}{sin x} \cdot \frac{sin x}{cos x}) - (sin x \cdot \frac{sin x}{cos x}) = (\frac{sin x}{sin x cos x}) - (\frac{sin x sin x}{cos x})$ $(1/sin x * sin x/cos x) - (sin x * sin x/cos x) = (sin x /(sin x cos x)) - ((sin x sin x)/cos x)$

And simplify

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

Using the Pythagorean identity ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x + cos^2x = 1$ , we know that $1 - {sin}^{2} x = {cos}^{2} x$ $1 - sin^2x = cos^2x$ , so substitute that in

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

And finally, simplify

$\frac{{cos}^{2} x}{cos x} = \frac{cos x}{1} = cos x$ $cos^2x/cos x = cos x/1 = cos x$

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Question #f68d0

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