another identity

[xc630]

Hi this following identity is giving me trouble.

1+Sin A -Cos 2A/ cos A + Sin2A = tan A

I know I have cos 2A and sin 2A to work with but it doesn't seem to be leading me anywhere. I would appreciate some help.

 

[Unco]

Go with that hunch.

To start off, there isn't much else to do with sin(2A) than to . . .

. . . then factorise the denominator.

 

[xc630]

I get this when I factorize the bottom:

1+ Sin A - Cos 2A/ cos A + 2SinACosA =tanA
1+ SinA - Cos 2A/ cosA (1+2SinA) =tan A

What should I do next?

 

[Mrspi]

I get this when I factorize the bottom:

1+ Sin A - Cos 2A/ cos A + 2SinACosA =tanA
1+ SinA - Cos 2A/ cosA (1+2SinA) =tan A

What should I do next?

For starters, some grouping symbols would be nice:
(1 + sin A - cos 2A) / [cos A(1 + 2 sin A)]

When I first looked at your problem, I did not know that the left side was a single fraction!!

Ok...now let's work on the numerator. In a previous post, you indicated that you knew several expressions for cos 2A. Let's use this one:
cos 2A = 1 - 2 sin<SUP>2</SUP> A

Substitute 1 - 2 sin<SUP>2</SUP> A for cos 2A:

[1 + sin A - (1 - 2 sin<SUP>2</SUP> A)] / [cos A (1 + 2 sin A)]
[1 + sin A - 1 + 2 sin<SUP>2</SUP> A] / [cos A (1 + 2 sin A)]

(sin A + 2 sin<SUP>2</SUP> A) / [cos A (1 + 2 sin A)]

Factor the numerator....reduce the fraction.....see what you have left.

I hope this helps you.