Question

How do I verify the identity of this?

tan(x) + cot(x) = 1 / (sin(x) * cos(x))

I know each side equals csc(x) * sec(x) but how do I get the answer?

@jim_thompson5910 Could you please help me?

Answer 1

All I know is that I can make the left side equal to
(sin(x) / cos(x)) + (cos(x) / sin(x))

Answer 2

Okay, why don't you make a common denominator on the left side?

Answer 3

Both sides need to have a denominator of sin(x) * cos(x)?

Answer 4

One thing I like to do when doing trig identities is to make a substitution like

c = cos (x)
s = sin (x)

just makes the writing a bit easier, then when I'm done with the algebra (like adding these two fractions) I undo the substitution. same with rationalizing radicals...

Answer 5

I meant both parts of the left.

Answer 6

So your problem would look like \[\frac{s}{c} + \frac{c}{s} = \frac{1}{c*s}\]

Answer 7

If nothing else, it's a lot easier to type into OS :-)

Answer 8

So to make a common denominator on the left side, what would you do?

Answer 9

(sin(x) / cos(x)) + (cos(x) / sin(x)) would become

(cos(x) * sin(x) / cos(x) * sin(x)) + (cos(x) * sin(x) / sin(x) * cos(x))

Sorry, I'm very rusty on my fraction skills.

Answer 10

You have \[\sin/\cos + \cos/\sin\] just get a common denominator and simplify the numerator. You'll see it when you get it in one fraction :)

Answer 11

Let's just do it with the letters, like I suggested:
\[\frac{s}{c} + \frac{c}{s} = \frac{1}{c*s}\]Multiply the top and bottom of the left fraction by the denominator of the right fraction:
\[\frac{s}{c}*\frac{s}{s} + \frac{c}{s} = \frac{1}{c * s}\]Now do the same for the right fraction:
\[\frac{s}{c}*\frac{s}{s} + \frac{c}{s}*\frac{c}{c}= \frac{1}{c * s}\]Combine the two fractions
\[\frac{s^2+c^2}{c*s} = \frac{1}{c * s}\]Now we can just eliminate the denominator
\[s^2+c^2=1\]And if we substitute back our trig functions, it becomes\[\sin^2 x+ \cos^2 x = 1\]And hopefully you recognize that very basic identity!

Answer 12

Thanks! I understand it now. I never thought it was that simple ha ha. Do you mind helping me with another question?