# Trig identity (sinx+cosx)^2tanx = tanx+2sin^2x [Solved!]

**Alexandra** 25 Dec 2015, 21:03

### My question

Show (sinx+cosx)^2tanx=tanx+2sin^2x

### Relevant page

1. Trigonometric Identities

### What I've done so far

`RHS = tanx+2sin^2x`

`=tan x + 2(1-cos^2x)`

`=tanx + 2 - 2cos^2 x`

But I can't make it look like the LHS.

X

Show (sinx+cosx)^2tanx=tanx+2sin^2x

Relevant page
<a href="/analytic-trigonometry/1-trigonometric-identities.php">1. Trigonometric Identities</a>
What I've done so far
`RHS = tanx+2sin^2x`
`=tan x + 2(1-cos^2x)`
`=tanx + 2 - 2cos^2 x`
But I can't make it look like the LHS.

## Re: Trig identity (sinx+cosx)^2tanx = tanx+2sin^2x

**Murray** 27 Dec 2015, 00:13

It's correct so far, but I don't think it helped.

Expand out the bracket on the LHS and see if you recognize anything.

X

It's correct so far, but I don't think it helped.
Expand out the bracket on the LHS and see if you recognize anything.

## Re: Trig identity (sinx+cosx)^2tanx = tanx+2sin^2x

**Alexandra** 27 Dec 2015, 21:48

OK.

` (sinx+cosx)^2tanx ` `= (sin^2x + 2sinx cosx + cos^2 x)tanx`

`=(1 + 2sinxcosx)tanx`

Is that right? But what do I do now?

X

OK.
` (sinx+cosx)^2tanx ` `= (sin^2x + 2sinx cosx + cos^2 x)tanx`
`=(1 + 2sinxcosx)tanx`
Is that right? But what do I do now?

## Re: Trig identity (sinx+cosx)^2tanx = tanx+2sin^2x

**Murray** 28 Dec 2015, 14:53

Just remember `tanx = (sinx)/(cosx)`

X

Just remember `tanx = (sinx)/(cosx)`

## Re: Trig identity (sinx+cosx)^2tanx = tanx+2sin^2x

**Alexandra** 29 Dec 2015, 05:44

`(1+2sinxcosx)tanx` `=(1+2sinxcosx)(sinx)/(cosx)`

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

`=tanx + 2sin^2 x`

`=RHS`

Thanks a lot.

X

`(1+2sinxcosx)tanx` `=(1+2sinxcosx)(sinx)/(cosx)`
`= (sinx)/(cosx) + 2sin^2x`
`=tanx + 2sin^2 x`
`=RHS`
Thanks a lot.

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