Question

Trig Question.

Answer 1

\[\frac{ \sec A - \tan A }{ \sec A + \tan A } = 1 - 2secAtanA + 2\tan^2 A\]

Answer 2

@campbell_st

Answer 3

What?

Answer 4

ok... @rebeccaxhawaii could do this one...

Answer 5

but start by simplifying the denominator

secA + tanA =

Answer 6

1/cos A + sinA/cosA

Answer 7

so what does it look like as 1 fraction...?

Answer 8

1+sinA all over cos A

Answer 9

great... so put that aside for the moment

do a similar thing with the numerator... what would you get..?

Answer 10

1-sinA all over cos A

Answer 11

ok... so if you rewrite the fraction you get

\[\frac{\frac{1 - sinA}{cosA}}{\frac{1 + sinA}{cosA}}\]

so I'd flip and multiply next..?

Answer 12

1-sinA/1+sinA

Answer 13

I just realised what I need to do...

scrub all that we are going to take the original equation and multiply numerator and denominator by (SecA - tanA)

because (secA -tanA)/(secA- tanA0 = 1

please trust me...

Answer 14

so you have

\[\frac{(secA - tanA)}{secA + tanA)} \times \frac{secA - tanA}{secA - tanA}\]

if you do that you get

\[\frac{(secA-tanA)^2}{\sec^2A - \tan^2A}\]

the denominator is an identity

Answer 15

thats 1

Answer 16

great so distribute the numerator

\[(secA - tanA)^2 = \sec^2A - 2secAtanA + \tan^2A\]

now you need an identity in terms of tan for \[\sec^2A = ?\]

Answer 17

tan^2 A +1

Answer 18

so make the substitution and then collect the like terms...

and sorry for taking a while to get the correct method

Answer 19

it gives the same solution... just takes a little longer

Answer 20

Thanks.

Answer 21

ask @rebeccaxhawaii to help with any more... i'm done..

Answer 22

hope it all helps... its just about using basic skills