`1/tanx+cosx=(2cosx)/sinx` is not identity.

First let's **rewrite left side by using only sine and cosine**:

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

Now solve equation `cosx/sinx+cosx=2cosx/sinx`

`cosx/sinx-cosx=0` Now multiply equation by `sinx.`

`cosx-cosxsinx=0`

`cosx(1-sinx)=0=>cosx=0` or `1-sinx=0`

`cosx=0=>x=pi/2+ k pi, k in ZZ`

`1-sinx=0`

`sinx=1=>x=pi/2+2k pi, k in ZZ`

This means that expression `1/tanx+cosx=(2cosx)/sinx` is true only for `x=pi/2+k pi, k in ZZ`, that is your expression is not identity because it's not true for all `x in RR.`

You should remember that `cot x = 1/tan x,` hence, you may substitute `cot x` for `1/tan x` such that:

`cot x + cos x = 2 cos x/sin x`

You also need to remember that `cos x/sin x = cot x` , hence, you may substitute `cot x` for `cos x/sin x` such that:

`cot x + cos x = 2 cot x => cos x = 2 cot x - cot x`

`cos x = cot x => cos x = cos x/sin x`

`cos x- cos x/sin x = 0`

Factoring out cos x yields:

`cos x(1 - 1/sin x) = 0 => cos x = 0 `

`1 - 1/sin x= 0 => 1/sin x = 1 => sin x = 1 `

**Hence, the given expression becomes an identity if `cos x = 0` or `sin x = 1` .**