trig identities help asap!

[Shooter1989]

ok i got 3 problems i cant solve and was wondering if anyone out there can help me out.. i cant really type the way the equation looks but try to follow me here! my teacher said that the answers is whats on the right side of the equation so everything on the left side needs to be converted to look like the right. here they are...

1. secx- 1/secx=sinxtanx
^ division sign

2. tanx(cotx-cosx)=1-sinx


3. sin(a+b)cosb-cos(a+b)sinb=sina


Thanks in advance!

 

[royhaas]

Start with the basic definitions and identities. For example,
secx-1/secx=1/cosx-cosx.

 

[Unco]

ok i got 3 problems i cant solve and was wondering if anyone out there can help me out.. i cant really type the way the equation looks but try to follow me here! my teacher said that the answers is whats on the right side of the equation so everything on the left side needs to be converted to look like the right. here they are...

When proving identities, we can manipulate either side (independently of the other) to shows LHS = RHS.

1. secx- 1/secx=sinxtanx

Being able to add/subtract fractions is a necessity.

\(\displaystyle \Huge \mbox{LHS = \sec{x} - \frac{1}{\sec{x}} = \frac{\sec^2{x} - 1}{\sec{x}}}\)

You should recognise the identity in the numerator to simplify. Put in terms of \(\displaystyle \mbox{\sin{x}}\) and \(\displaystyle \mbox{\cos{x}}\) to simplify further and show this equals the RHS.

2. tanx(cotx-cosx)=1-sinx

Expand the LHS. Simplify, putting in terms of \(\displaystyle \mbox{\sin{x}}\) and \(\displaystyle \mbox{\cos{x}}\) should make things clear.

3. sin(a+b)cosb-cos(a+b)sinb=sina

Compare the LHS with the compound angle formula for \(\displaystyle \mbox{\sin{(\alpha -\beta)}}\).

Edit: minus.