Trouble With Trig Proofs: Seeking Help

[seiferseph]

I recently posted about some trig equations, now I'm doing some HW on trig proofs, i got the first couple trig proofs, but had trouble with the last two.
Here are the two problems (attached). For the first one, i can't even get started. i have some ideas, but i can't find out how to get the right side (1/(1+secθ). For the second one, i simplify the right side for cosθ/sinθ, but I'm not sure if that's right (i can't get the right side to anything like that). any help would be appreciated, thanks!
http://i2.photobucket.com/albums/y15/seiferseph/untitled.jpg

 

[Jameson]

I can't access the attachment you posted...

 

[seiferseph]

I can't access the attachment you posted...

it says its pending approval, is there anywhere else i can upload it?
does this work
http://i2.photobucket.com/albums/y15/seiferseph/untitled.jpg

 

[HallsofIvy]

It's open now- put into latex it says:
$$1. \frac{1-sin^2(\theta)-2cos(\theta)}{cos^2(\theta)-cos(\theta)-2}= \frac{1}{1+ sec(\theta)}$$

$$2. \frac{cos(\theta)-sin(\theta)}{cos(\theta)+sin(\theta)}=sec(2\theta)- tan(2\theta)$$


Generally speaking, my first step with trig identities is to convert everything into sine and cosine. Here, sec(θ)= 1/cos(θ) so I think that I would in fact convert every thing to cos(θ)- sin(θ) only appears as sin²(θ) in problem 1 so that is just
$$\frac{cos^2(\theta)- 2cos(\theta)}{cos^2(\theta)-cos(\theta)-2}= \frac{1}{1+\frac{1}{cos(\theta)}}= \frac{cos(\theta)}{cos(\theta)+1}$$
Now think about factoring the terms on the left.

 

[seiferseph]

thank you, i came close, i had the right side simplified and all (and had the bottom on the left factored), but i couldn't get the top on the left. thanks! now for the 2nd one, is it simplified down to cosθ/sinθ?

 

[Data]

Neither side simplifiees to $$\frac{\cos{\theta}}{\sin{\theta}}$$. You will need to remember what the conjugate of $$a-b$$ is, if you do it the same way I did. The conjugate of $$a-b$$ is $$a+b$$. Recall that

$$\frac{a+b}{a-b} = \frac{(a+b)^2}{(a-b)(a+b)} = \frac{(a+b)^2}{a^2-b^2}$$

You might be needing it :)

 

[seiferseph]

how does the conjugate work? i know you can use it if they are on opposite sides (and diagonal), but how do you use it here. i did multiply the top by (cosθ+sinθ) and then make the bottom (cosθ+sinθ)^2, is that correct? because i solved, but got the reciprocal of the right side.

 

[Data]

You've made a mistake then. Try multiplying top and bottom of the left side by $$\cos{\theta} - \sin{\theta}$$ instead, so that the denominator is a difference of squares. You will then need to make several simplifications using identities.

The way that the conjugate "works" is just multiplication by 1:

$$\frac{a-b}{a+b} = \frac{a-b}{a+b} \cdot \frac{a-b}{a-b} = \frac{(a-b)^2}{a^2-b^2}$$

 

[seiferseph]

You've made a mistake then. Try multiplying top and bottom of the left side by $$\cos{\theta} - \sin{\theta}$$ instead, so that the denominator is a difference of squares. You will then need to make several simplifications using identities.

The way that the conjugate "works" is just multiplication by 1:

$$\frac{a-b}{a+b} = \frac{a-b}{a+b} \cdot \frac{a-b}{a-b} = \frac{(a-b)^2}{a^2-b^2}$$

ok, so you can multiply both top and bottom by either (cosθ + sinθ) or (cosθ - sinθ) to simplify one into the difference of squares?
thanks! i solved it, getting (1-sin2θ)/(cos2θ) on both sides