Two triq Identites?

Question

uhgggg ok this is so simple, i dont see it

1+sin(x)/cos(x) = cos(x)/1-sin (x)

is there an id for 1+sin =???

also

cos^2(x) - 3cos(x)+2 / sin^2(x) = 2-cos(x) / 1+cos(x)

no clue where to start on that one..

thanks

Answer 1

First one:

Multiply RHS by (1+Sin x)/(1+Sin x) to get:

[Cos x (1+sin x)]/[1-Sin^2(x)]

But 1-Sin^2(x) = Cos^2(x), so we have:

Cos x (1+sin x)/Cos^2(x)

Now divide top and bottom by Cos x to get LHS.

Answer 2

cross multiply the two sides of the equal sign.
You get 1-sin^2(x) = cos^2(x)

bring the sin^2(x) over and you get 1=sin^2(x) + cos^2(x)

and using your trig identities that solves to 1=1 (i think, its been a while since I've done this)

Answer 3

1. The way you wrote this leaves the fractions a little ambiguous, but I'll assume that this is
(1 + sin(x)) / cos(x) = cos(x) / (1-sin(x))
Multiply both sides by cos(x), then by (1-sin(x)):
1 + sin(x) = cos²(x) / (1-sin(x))
[ 1 + sin(x) ][ 1-sin(x) ] = cos²(x)
1 + sin(x) - sin(x) - sin²(x) = cos²(x)
1 = sin²(x) + cos²(x)
This is a well-known trig identity.

2. Again, I'm going to have to make some assumptions here about your fractions:
[cos²(x) - 3cos(x)+2] / sin²(x) = [2-cos(x)] / [1+cos(x)]
[cos²(x) - 3cos(x)+2] = sin²(x)[2-cos(x)] / [1+cos(x)]
[cos²(x) - 3cos(x)+2] = (1-cos²(x))[2-cos(x)] / [1+cos(x)]
[cos²(x) - 3cos(x)+2] = (1+cos(x)) [1-cos(x)] [2-cos(x)] / [1+cos(x)]
[cos²(x) - 3cos(x)+2] = [1-cos(x)] [2-cos(x)]
cos²(x) - 3cos(x)+2 = 2 - 2cos(x) - cos(x) + cos²(x)
cos²(x) - 3cos(x) = - 2cos(x) - cos(x) + cos²(x)
-3cos(x) = - 2cos(x) - cos(x)
-3cos(x) = -3cos(x)