# Quick and easy question trig identities

• Jan 12th 2008, 02:38 PM
Slipery
Quick and easy question trig identities
Hey, I am just starting my trig identities unit and I have a question I'm stuck on.. it's probably very easy, so bear with me and don't make fun of me ^^

(sec˛x + csc ˛x) - (tan˛x - cot˛x)
So I have ( 1/cos˛x+ 1/sin˛x) - (sin˛x/cos˛x+ cos˛x/sin˛x)...
• Jan 12th 2008, 02:44 PM
Jhevon
Quote:

Originally Posted by Slipery
Hey, I am just starting my trig identities unit and I have a question I'm stuck on.. it's probably very easy, so bear with me and don't make fun of me ^^

(sec˛x + csc ˛x) - (tan˛x - cot˛x)
So I have ( 1/cos˛x+ 1/sin˛x) - (sin˛x/cos˛x+ cos˛x/sin˛x)...

i suppose you want to simplify here? that's good. now just combined the fractions in each set of brackets. then combine the two resulting fractions
• Jan 12th 2008, 03:06 PM
Slipery
Yes, just simplify.

For the first bracket, (1/cos˛x + 1/sin˛x)
can I change the cos˛x and sin˛x to 1 and have it as 1+1= 2?
• Jan 12th 2008, 03:20 PM
Soroban
Hello, Slipery!

Your work is correct, but are you aware of these identities?

. . . $\displaystyle \begin{array}{ccc}\sec^2\!x \:=\:\tan^2\!x + 1 & \Rightarrow & \sec^2\!x - \tan^2\!x \:=\:1 \\ \csc^2\!x \:=\:\cot^2\!x + 1 & \Rightarrow & \csc^2\!x - \cot^2\!x \:=\:1\end{array}$

Quote:

Simplify: .$\displaystyle (\sec^2\!x + \csc^2\!x) - (\tan^2\!x + \cot^2\!x)$

We have: .$\displaystyle (\sec^2\!x - \tan^2\!x) + (\csc^2\!x - \cot^2\!x) \;\;=\;\;1 + 1 \;\;=\;\;2$

• Jan 12th 2008, 03:37 PM
Slipery
Ahhh, that is very helpful. thank you both soroban and jhevon.
I honestly think my brain has issues with this type of work, it just wont' draw the connections.