1. ## trig identity prob

-Simplify each expression using the fundamental identities

2) Sec theta x cos theta

3) 1/csc^2x + 1/sec^2x

4) 1-sin^2u/cosu

5) cos^2x+ (sin x +1)^2/sin x + 1

Pleeease help, I don't know what to do
thanks!

2. Originally Posted by zerostumbleine33
-Simplify each expression using the fundamental identities

2) Sec theta x cos theta

3) 1/csc^2x + 1/sec^2x

4) 1-sin^2u/cosu

5) cos^2x+ (sin x +1)^2/sin x + 1

Pleeease help, I don't know what to do
thanks!
here's number 2!

$sec \theta cos\theta$

$sec \theta = \frac{1}{cos\theta}$

$\frac{1}{cos\theta}*\frac{cos\theta}{1}=1$

3. Originally Posted by zerostumbleine33
-Simplify each expression using the fundamental identities

2) Sec theta x cos theta

3) 1/csc^2x + 1/sec^2x

4) 1-sin^2u/cosu

5) cos^2x+ (sin x +1)^2/sin x + 1

Pleeease help, I don't know what to do
thanks!
Please use parenthesis! I presume that 4 is to read $\frac{1 - sin^2(u)}{cos(u)}$ and not as $1 - \frac{sin^2(u)}{cos(u)}$.

My first rule of thumb is to get everything in terms of sine and cosine. This will immediately solve 2 and 3.

Also look for $sin^2(u) + cos^2(u) = 1$ and
$sin(2u) = 2sin(u)cos(u)$ and $cos(2u) = 1 - 2sin^2(u) = 2cos^2(u) - 1 = cos^2(u) - sin^2(u)$.

Expand any expressions under exponents. ie. $(sin(x) + 1)^2 = sin^2(x) + 2sin(x) + 1$.

Finally simplify and see if you can factor anything after you have done the above steps.

-Dan

4. Hello, zerostumbleine33!

It looks like you don't know any of the basic identities . . . *sigh*

Simplify each expression using the fundamental identities

$3)\;\;\frac{1}{\csc^2\!x} + \frac{1}{\sec^2\!x}$

We have: . $\left(\frac{1}{\csc x}\right)^2 + \left(\frac{1}{\sec x}\right)^2$

You're expected to know that: . $\frac{1}{\csc x} \:=\:\sin x$ .and . $\frac{1}{\sec x} \:=\:\cos x$

$\text{So we have: }\;(\sin x)^2 + (\cos x)^2 \;=\;\underbrace{\sin^2\!x + \cos^2\!x)}_{\text{This is 1, right?}}$

Final answer: . $1$

$4)\;\;\frac{1 - \sin^2\!u}{\cos u}$

You're supposed to know that: . $\sin^2\!u + \cos^2\!u \:=\:1$

. . and you should recognize: . $1 - \sin^2\!u \:=\:\cos^2\!u$

So we have: . $\frac{1-\sin^2\!u}{\cos u} \;=\;\frac{\cos^2\!u}{\cos u} \;=\;\cos u$