Trig Identity

Just reaarange:

(1-sinx)/cosx = cosx/(1+sinx) (i would have used equivalent sign but there isn't one)

=> (1+sinx)(1-sinx) = (cosx)^2
=> 1 - (sinx)^2 = (cosx)^2

Which using sin^2 + cos^2 = 1 shows it's true.

Is that cross multiplying ? Not sure how its meant to be rearranged ?

Thanks =)

Yeah, it's just rearranging equal fractions

Like if a/b = c/d then ad = bc

I got stuck on it on the part now that

I end up with:

1-sin^2 theta = cos^2theta SAME AS cos theta over 1+sintheta

Not sure what to do now, I can move the - sin squared theta to get 1 on the left hand side, but not sure how it equals the other side ?

u have 1 - sin^2 = cos^2?

If you have that, then that proves it.

You can plug in a value of theta if needed just to verify it.

Thanks alot, i somehow made the beginning one come back in my working out, all fine now. Can you help on the 2nd aswell please ?

Just reaarange:

(1-sinx)/cosx = cosx/(1+sinx) (i would have used equivalent sign but there isn't one)

=> (1+sinx)(1-sinx) = (cosx)^2
=> 1 - (sinx)^2 = (cosx)^2

Which using sin^2 + cos^2 = 1 shows it's true.

listen to dnkt.

If you're doing C3 trig identities, you have to make the LHS equal the RHS through manipulation, or RHS to LHS.

Actually you don't. Although it is certainly considered best practice to work on one side, full marks will still be awarded if both sides are manipulated.