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

**Alexandra** 29 Dec 2015, 18:45

### 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

**Murray** 30 Dec 2015, 03:12

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

**Alexandra** 31 Dec 2015, 04:08

`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

**Murray** 31 Dec 2015, 20:26

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

**Alexandra** 01 Jan 2016, 20:13

`(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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