if theta is equal to pi/4 , beta is equal to pi/2, sec theta - beta = ? answers: a) - csc pi/4 b) sec pi/4 c) csc pi/4 d) - sec pi/4
So lets say \[\huge\color{blue}{ θ = \frac{π}{4}} \]\[\huge\color{blue}{ β = \frac{π}{2}} \]
\[\huge\color{red}{ π=180°}\]
\[ \huge\color{blue}{ Sec~(θ-β)} ~~~~~~~~~~~this?\]
I would simply do this the following way. \[ \huge\color{blue}{ \sec(θ-β) ~⇒ ~\sec(\frac{π}{4}-\frac{π}{2})~⇒} \]\[ \huge\color{blue}{ \sec(\frac{π}{4}-\frac{2π}{4})~⇒} \huge\color{blue}{ \sec(-\frac{π}{4})}\]CAN YOU GO FROM HERE?
Yes, thank you very much
Anytime, I would probably be able to do this without trig.
hey, what is the next step?
Should I say the next step, @lucaz?
I don't know ho to go on
how*
\[\sec(-π/4)=-\sec(π/4)\]\[\sec(π/4)=\sqrt{2}\]So,\[-\sec(π/4)=-\sqrt{2}\] @lucaz, please don't tell me I don't know how to go on, I got a 97 on trig last year.
@Lucaz, should I prove that \[\sec(π/4)=\sqrt{2}\]or your good?
@lucaz why are we so silent?
no, it's ok
Oh, tnx a lot lucaz. I am not good at math but this staff I can do.
lol, I wasn't testing you
Oh, @lucaz I was blind
I thought you said YOU don't know how to go on.
xD
SORRY @lucaz apologize!!
right =]
Anyway, identities used: \[\sin(-x)=-\sin(x)\]\[π=180\]\[Trigonometric~~~table.\] adding fractions -prealgebra.
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