Question #dd613

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Question #dd613

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Answer 1 · 2017-12-15T01:02:00

$sin x$ $sinx$

Explanation:

First simplify the inverse trig identities.
$csc x = \frac{1}{sin x}$ $cscx=1/sinx$ ; $cot x = \frac{cos x}{sin x}$ $cotx=cosx/sinx$
Plugging those in you get
$\frac{1}{sin x} - cos x (\frac{cos x}{sin x})$ $1/sinx-cosx(cosx/sinx)$ multiply
$\frac{1}{sin x} - \frac{{cos}^{2} x}{sin x}$ $1/sinx-cos^2x/sinx$ combine them together
$\frac{1 - {cos}^{2} x}{sin x}$ $(1-cos^2x)/sinx$ Use trig identities of squared trig functions
${sin}^{2} x + {cos}^{2} x = 1 \Rightarrow {sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x+cos^2x=1 => sin^2x=1-cos^2x$
$\frac{{sin}^{2} x}{sin x} \Rightarrow sin x$ $sin^2x/sinx=>sinx$ Reduce

Answer 2 · 2017-12-15T01:02:06

$csc x - cos x \cdot cot x = sin x$ $cscx-cosx*cotx=sinx$

Explanation:

$csc x - cos x \cdot cot x$ $cscx-cosx*cotx$

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

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

= $\frac{{(sin x)}^{2}}{sin x}$ $(sinx)^2/sinx$

= $sin x$ $sinx$

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Question #dd613

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