How do I prove that $cscx-sinx=cosxcotx$ ?

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How do I prove that $cscx-sinx=cosxcotx$ ?

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Answer 1 · 2017-05-19T14:49:47

$cscx-sinx$

$=1/sinx-sinx$

$=(1-sin^2x)/sinx$

$=cos^2x/sinx$

$=cosx/sinx xxcosx$

$=cotxcosx$

$thereforecscx-sinx=cotxcosx$ $sf(QED)$

Answer 2 · 2017-05-19T14:50:39

See proof below

Explanation:

We need

$cscx=1/sinx$

$sin^2x+cos^2x=1$

$cotx=cosx/sinx$

Therefore,

$LHS=cscx-sinx$

$=1/sinx-sinx$

$=(1-sin^2x)/sinx$

$=cos^2x/sinx$

$=cosx*cosx/sinx$

$=cosxcotx$

$=RHS$

$QED$

Answer 3 · 2017-05-19T15:00:31

We have: $csc(x) - sin(x)$

Let's apply a standard trigonometric identity; $csc(x) = frac(1)(sin(x))$ :

$= frac(1)(sin(x)) - sin(x)$

$= frac(1)(sin(x)) - frac(sin^(2)(x))(sin(x))$

$= frac(1 - sin^(2)(x))(sin(x))$

One of the Pythagorean identities is $cos^(2)(x) + sin^(2)(x) = 1$

We can rearrange it to get:

$Rightarrow cos^(2)(x) = 1 - sin^(2)(x)$

Let's apply this rearranged identity to our proof:

$= frac(cos^(2)(x))(sin(x))$

$= cos(x) times frac(cos(x))(sin(x))$

Finally, let's apply another standard trigonometric identity; $cot(x) = frac(cos(x))(sin(x))$ :

$= cos(x) times cot(x)$

$= cos(x) cot(x)$ Q.E.D.

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How do I prove that $cscx-sinx=cosxcotx$ ?

3 Answers

Explanation:

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