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Prove the trig identity cosx/(secx+tanx)= 1-sinx [Solved!]

My question

Prove the trig identity cosx/(secx+tanx)= 1-sinx

Relevant page

1. Trigonometric Identities

What I've done so far

The LHS looks the most complicated, but what do you do with it?

X

Prove the trig identity cosx/(secx+tanx)= 1-sinx
Relevant page

<a href="/analytic-trigonometry/1-trigonometric-identities.php">1. Trigonometric Identities</a>

What I've done so far

The LHS looks the most complicated, but what do you do with it?

Re: Prove the trig identity cosx/(secx+tanx)= 1-sinx

When you see a mix of trigonometric ratios, try converting everything to just sin, cos and/or tan.

X

When you see a mix of trigonometric ratios, try converting everything to just sin, cos and/or tan.

Re: Prove the trig identity cosx/(secx+tanx)= 1-sinx

`LHS = cosx/(secx+tanx)`

`=(cos x)/(1/(cosx) + (sinx)/(cosx))`

I multiply top and bottom by `cosx`:

`=(cos^2 x)/(1 + sinx)`

It doesn't look like the RHS yet

X

`LHS = cosx/(secx+tanx)`

`=(cos x)/(1/(cosx) + (sinx)/(cosx))`

I multiply top and bottom by `cosx`:

`=(cos^2 x)/(1 + sinx)`

It doesn't look like the RHS yet

Re: Prove the trig identity cosx/(secx+tanx)= 1-sinx

OK, good so far.

This uses a trick that you can see on the page you originally came from:

1. Trigonometric Identities

You need to multiply top and bottom by `1-sinx`.

Why? Because (experience tells me) it will help us get it in the right form.

X

OK, good so far.

This uses a trick that you can see on the page you originally came from:

<a href="/analytic-trigonometry/1-trigonometric-identities.php">1. Trigonometric Identities</a>

You need to multiply top and bottom by `1-sinx`. 

Why? Because (experience tells me) it will help us get it in the right form.

Re: Prove the trig identity cosx/(secx+tanx)= 1-sinx

`(cos^2x (1-sinx))/((1+sinx)(1-sinx))`

`=(cos^2x (1-sinx))/(1-sin^2x)`

`=(cos^2x (1-sinx))/(cos^2x)`

`=1-sinx`

`=RHS`

I think I'm starting to get it!

X

`(cos^2x (1-sinx))/((1+sinx)(1-sinx))`

`=(cos^2x (1-sinx))/(1-sin^2x)`

`=(cos^2x (1-sinx))/(cos^2x)`

`=1-sinx`

`=RHS`

I think I'm starting to get it!

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