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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!