How do you prove $\frac{sin x + cos x cot x}{cot x} = sec x$ $(sinx + cosxcotx)/cotx = secx$ ?

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Answer 1 · 2015-05-17T12:12:45

We need to get the left side equal to the right side:

So we will look at $= \frac{sin x + cos x cot x}{cot x}$ $=(sinx + cosxcotx)/cotx$

we can use the identity of $cot x = \frac{cos x}{sin x}$ $color(red)(cotx = cosx/sinx)$ here

so we will have $= \frac{sin x + cos x (\frac{cos x}{sin x})}{\frac{cos x}{sin x}}$ $=(sinx + cosx(cosx/sinx))/(cosx/sinx)$

if we now take out a $\frac{1}{sin x}$ $color(red)(1/sinx)$ from the top, we will have:

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

there we can use the identity that ${sin}^{2} x + {cos}^{2} x = 1$ $color(red)(sin^2x + cos^2x = 1)$

thus we are left with $= \frac{\frac{1}{sin x}}{\frac{cos x}{sin x}}$ $=(1/sinx)/(cosx/sinx)$

which we can write as $= (\frac{1}{sin x}) (\frac{sin x}{cos x})$ $=(1/sinx)(sinx/cosx)$

the $sin x$ $color(red)(sinx)$ 's will cancel each other, and we will be left with:

$= (\frac{1}{cos x})$ $= (1/cosx)$

which is our identity to $(\frac{1}{cos x}) = sec x$ $color(red)((1/cosx) = secx)$

thus we get $= sec x$ $=secx$

and we have that the right hand side $=$ $=$ to left hand side

How do you prove sinx+cosxcotxcotx=secx(sinx + cosxcotx)/cotx = secx?