#1**+23 **

Verify cot(x) * sec(x) = csc(x)

cot(x) = 1/tan(x)

tan(x) = sin(x)/cos(x)

So cot(x) = 1/ sin(x) * cos(x)

And sec(x) = 1/cos(x)

So the whole equation is:

1/ sin(x) * cos(x) * 1/cos(x) =csc(x)

The cos(x) and 1/cos(x) cancel out:

1/sin(x) = csc(x)

csc(x) is the same as 1/sin(x):

csc(x) = csc(x)

:)

kitty<3
Jul 16, 2014

#1**+23 **

Best Answer

Verify cot(x) * sec(x) = csc(x)

cot(x) = 1/tan(x)

tan(x) = sin(x)/cos(x)

So cot(x) = 1/ sin(x) * cos(x)

And sec(x) = 1/cos(x)

So the whole equation is:

1/ sin(x) * cos(x) * 1/cos(x) =csc(x)

The cos(x) and 1/cos(x) cancel out:

1/sin(x) = csc(x)

csc(x) is the same as 1/sin(x):

csc(x) = csc(x)

:)

kitty<3
Jul 16, 2014

#2**+8 **

Thanks Kitty,

Kitty has done it the usual traditional way (the way that Iwould normally do it) - I just thought I would take a look at a more basic method.

$$\\cot\;\theta=\frac{1}{tan\theta}=\frac{adj}{opp}\qquad \mbox{pos in 1st and 3rd quads}\\\\

sec\;\theta=\frac{1}{cos\theta}=\frac{hyp}{adj}\qquad \mbox{pos in 1st and 4th quads}\\\\

cosec\;\theta=\frac{1}{sin\theta}=\frac{hyp}{opp}\qquad \mbox{pos in 1st and 2nd quads}\\\\

\cot\theta\times sec\;\theta=\frac{adj}{opp}\times\frac{hyp}{adj}=\frac{hyp}{opp}=cosec\;\theta$$

Now, this obviously works in the first quadrant where everything is positive but what about in the other quadrants?

2nd quad - x - = + true

3rd quad + x - = - true

4th quad - x + = - true

Melody
Jul 17, 2014