Question

Simplify the expression and write in terms of sine.

tanØ secØ

Answer 1

tan(x)*sec(x)=sin(x)/cos(x)*1/cos(x)=sin(x)/cos^2(x)=sin(x)/1-sin^2(x)

Answer 2

Note that \(\sec\theta = \dfrac{1}{\cos\theta}\), so this problem is similiar to your other one here: http://openstudy.com/study#/updates/54034070e4b0f2ed1e13ce94

Answer 3

Isn't \[\sec^2\theta= \frac{ 1 }{ \cos^2\theta } \] that what my teacher showed me. @chad123 @micahwood50

Answer 4

yep thats right

Answer 5

tanØ × secØ = sinØ × secØ × secØ = sinØ sec²Ø

Answer 6

Don't I have to turn \[\sec^2\theta \] to sine? @idku

Answer 7

\[\tan \theta = \frac{ \sin \theta }{ \cos \theta }\] and \[\sec \theta = \frac{ 1 }{ \cos \theta }\] so isn't it \[ \left( \frac{ \sin \theta }{ \cos \theta } \right)\left( \frac{ 1 }{ \cos \theta } \right)\] @idku @chad123 @micahwood50

Answer 8

Yeah, hence you have \(\dfrac{\sin\theta}{\cos^2\theta}\)

Answer 9

I am sure you can rewrite denominator in terms of sine

Answer 10

\[\cos^2 \theta = \frac{ 1 }{ \sin^2 \theta }\] so... \[\frac{ \sin \theta }{ \sin^2 \theta }\] @micahwood50

Answer 11

Not right. \(\cos^2\theta \neq\dfrac{1}{\sin^2\theta}\)

Try to use this identity: \(\sin^2\theta+\cos^2\theta=1\)

Answer 12

\[\frac{ \sin \theta }{ 1-\sin^2 \theta }\] @micahwood50

Answer 13

Now that's right. :)