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Show (1-sinx)/(1+sinx)= (tanx-secx)^2 [Solved!]

My question

How to show `(1-sinx)/(1+sinx)= (tanx-secx)^2`?

Relevant page

1. Trigonometric Identities

What I've done so far

The RHS looks like:

`(tanx-secx)^2`

`=tan^2x - 2tanxsecx + sec^2x`

But I'm stuck again

X

How to show `(1-sinx)/(1+sinx)= (tanx-secx)^2`?
Relevant page

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

What I've done so far

The RHS looks like:

`(tanx-secx)^2`

`=tan^2x - 2tanxsecx + sec^2x`

But I'm stuck again

Re: Show (1-sinx)/(1+sinx)= (tanx-secx)^2

Remember the hint from before? Get everything in terms of sin and cos.

It makes it easier to recognize things!

X

Remember the hint from before? Get everything in terms of sin and cos.

It makes it easier to recognize things!

Re: Show (1-sinx)/(1+sinx)= (tanx-secx)^2

`tan^2x - 2tanxsecx + sec^2x`

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

It didn't seem to help

X

`tan^2x - 2tanxsecx + sec^2x`

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

It didn't seem to help

Re: Show (1-sinx)/(1+sinx)= (tanx-secx)^2

Have a look at the middle term. You are missing something...

X

Have a look at the middle term. You are missing something...

Re: Show (1-sinx)/(1+sinx)= (tanx-secx)^2

First, I would like to thank you for the help you are providing me.

Oops.

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

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

I guess LHS uses the same trick as before

`LHS = (1-sinx)/(1+sinx)xx (1-sinx)/(1-sinx)`

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

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

`=RHS`

Thanks a lot!

X

First, I would like to thank you for the help you are providing me.

Oops.

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

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

I guess LHS uses the same trick as before

`LHS = (1-sinx)/(1+sinx)xx (1-sinx)/(1-sinx)`

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

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

`=RHS`

Thanks a lot!

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