Hi,
I am doing my best to understand my trig class, but I just cannot understand proving identities. I've spent the last three night searching the internet for videos and tutorials and still do not understand my own problems...

I was able to figure out a solution to one of them by looking at a solution to a previously posted problem, but I can't get it to work for any of my others.

I really would like to understand these!

Problem 1:
$
\frac{cos\Theta + sin\Theta}{sec\Theta + csc\Theta} = (sin\Theta)(cos\Theta)
$

I started with LHS. I multiplied by $(sec\Theta - csc\Theta)$ on both numerator and denominator and got:

$
\frac{sin\Theta}{cos\Theta} - \frac{cos\Theta}{sin\Theta}$

over

$sec^2\Theta - csc^2\Theta$

And there I've lost it. Any hints? Am I even on the right track?

2. Originally Posted by Blue

Hi,
I am doing my best to understand my trig class, but I just cannot understand proving identities. I've spent the last three night searching the internet for videos and tutorials and still do not understand my own problems...

I was able to figure out a solution to one of them by looking at a solution to a previously posted problem, but I can't get it to work for any of my others.

I really would like to understand these!

Problem 1:
$
\frac{cos\Theta + sin\Theta}{sec\Theta + csc\Theta} = (sin\Theta)(cos\Theta)
$

I started with LHS. I multiplied by $(sec\Theta - csc\Theta)$ on both numerator and denominator and got:

$
\frac{sin\Theta}{cos\Theta} - \frac{cos\Theta}{sin\Theta}$

over

$sec^2\Theta - csc^2\Theta$

And there I've lost it. Any hints? Am I even on the right track?
This is a different approach:

$\dfrac{\cos(\theta) + \sin( \theta)}{\sec(\theta) + \csc(\theta)} = \dfrac{\cos(\theta) + \sin( \theta)}{\dfrac1{\cos(\theta)} + \dfrac1{\sin(\theta)}} = \dfrac{\cos(\theta) + \sin( \theta)}{\dfrac{\sin(\theta)+\cos(\theta)}{\sin(\t heta) \cdot \cos(\theta)}} = \sin(\theta) \cdot \cos(\theta)$

3. Originally Posted by Blue

I really would like to understand these!

Problem 1:
$
\frac{cos\Theta + sin\Theta}{sec\Theta + csc\Theta} = (sin\Theta)(cos\Theta)
$

I started with LHS. I multiplied by $(sec\Theta - csc\Theta)$ on both numerator and denominator and got:

$
\frac{sin\Theta}{cos\Theta} - \frac{cos\Theta}{sin\Theta}$

over

$sec^2\Theta - csc^2\Theta$

And there I've lost it. Any hints? Am I even on the right track?
Not a bad start at all. When you've got a binomial in the denominator, multiplying by the conjugate is always something to think about doing. One hint here, is that WHENEVER you have a squared trig function, think about using a Pythagorean identity. That would lead to putting in some $tan^2\theta$s and some $cot^2\theta$s in your denominator. I don't see how that would help though, because the denominator would look very similar to what you've already got there with the $sec^2\theta$s and $csc^2\theta$s. At that point, I'd say punt - which means start over and follow the idea below.

When you can't see ANYTHING else to do, an idea that can frequently be very useful (but can also lead to lots of work) is to change everything in the equation to $sin\theta$ and $cos\theta$. For this problem, that would lead to $\frac{cos\theta + sin\theta}{\frac{1}{cos\theta}+\frac{1}{sin\theta} }$. You can simplify this algebraically.

4. Thank you both! I can't believe I didn't see that before.