Gah: Trigonometry

I've been doing some trigonomtry homework and I've nearly got it done.  All I need done are these three problems... but I can't figure it out...

HAAAAAAAAAAALP!

I need to be able to solve the problems and understand them so working it out is appreciated.

1) 1+sin(theta)/cos(theta) + cos(theta)/1+sin(theta) = 2sec(theta)
2) sin(theta)/1-cot(theta) - cos(theta)/tan(theta)-1 = sin(theta) + cos(theta)
3) (3cos(theta) - 4sin(theta))(squared) + (4cos(theta) + 3cos(theta))(squared) = 25

Yeah you need a list of trigonmetric identities.  Stuff like sin#/cos# = tan# and sin²# +  cos²# = 1.

I do know my trigonometric identities.  Problem is, this is a proof and I can't figure these 3 out.  These are the hardest ones on my homework.

>>3
OK, (1).
a)combine the two fractions into a single fraction whose denominator is the product of the two denominators.  Each numerator gets muliplied by the other fraction's denominator in the process:  [(1 + sin#)² + cos²#]/[cos#(1 + sin#)]
b)multipy out the (1 + sin#)² to get 1 + 2sin# + sin²#.
c)trade the sin²# + cos²# for a 1, according to the basic pythagorean identity.
d) factor 2 out of each term in the numerator.
e) cancel (1 + sin#) off the numerator and denominator.
f) you are left with 2/cos#, which by definition of sec# = 1/cos#, is the same as 2sec#, matching the right side of the equation.

3. [3cos(θ) - 4sin(θ)]²+[4cos(θ) + 3cos(θ)]² != 25

3. [3cos(θ) - 4sin(θ)]² + [4cos(θ) + 3sin(θ)]² = 25

Number three's easy.
1. Expand
2. Cancel
3. Factor
4. sin²(θ) + cos²(θ) = 1

Thanks for solving the first one, Anonymous.

>>7
Could someone please elaborate on this?

[3cos(θ) - 4sin(θ)]² + [4cos(θ) + 3cos(θ)]² != 25

[3cos(θ) - 4sin(θ)]² + [4cos(θ) + 3sin(θ)]² = 25

1. Expand
[indent]9cos²(θ) - 24sin(θ)cos(θ) + 16sin²(θ) + 16cos²(θ) - 24sin(θ)cos(θ) + 9sin²(θ) = 25[/indent]
2. Cancel
[indent]9cos²(θ) + 16sin²(θ) + 16cos²(θ) + 9sin²(θ) = 25[/indent]
3. Factor
[indent]9[sin²(θ) + cos²(θ)] + 16[sin²(θ) + cos²(θ)] = 25[/indent]
4. sin²(θ) + cos²(θ) = 1
[indent]9 + 16 = 25[/indent]

>>9
Change one -24sin(θ)cos(θ) to +24sin(θ)cos(θ) in step 1. Expand.

>>10
Ah, I thought that looked a little funny.  Thanks Anonymous.

>>11
No problem. You can see 3 doesn't work just by letting θ equal a number.

(2) can be solved by first combining the two fractions into a single fraction.  A nice efficient way is to multiply the first fraction by tanθ/tanθ, because that will cause the denominator to match the second fraction's denominator.

The numerator of the combined fraction will have two terms: sinθtanθ - cosθ.  If you factor cosθ out of those, you will produce cosθ(sinθtanθ/cosθ - cosθ/cosθ) which becomes cosθ(tan²θ - 1).  The bracketed factor is a difference of squares which can be factored as (tanθ + 1)(tanθ - 1), allowing you to cancel a factor to eliminate the denominator.

The remaining factors multiply out to give the right side of the equation.