Question

Question #40916

Answer 1

Proof via considering trig identities

Explanation:

First we consier `1-2sin^2x` as being `cos^2x-sin^2x`
As; `1-2sin^2x` = `(1- sin^2x) -sin^2x`
and `1-sin^2x` = `cos^2x`

So we can simplify the LHS to `(cos^2x-sin^2x)/(cosx+sinx)`

Hence `cos^2x-sin^2x` can be written as `(cosx+sinx)(cosx-sinx)` by difference in two squares

Hence we can simplify the LHS to `((cosx+sinx)(cosx-sinx))/(cosx+sinx)`

Hence cancelling, to yields, `cosx-sinx`

`LHS = RHS`

Answer 2

See other answer.

Explanation:

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