Verify cos(x)+1/tan^2(x)=cos(x)/sec(x)-1 Using one side

[posenkid88]

Verify cos(x)+1/tan^2(x)=cos(x)/sec(x)-1 Using one side

I got as far as cos(x)+1/sec^2(x)-1 on the left side. Please help me!!! Thanks

 

[stapel]

Verify cos(x)+1/tan^2(x)=cos(x)/sec(x)-1 Using one side

Is the left-hand side one of the following?

. . . . .cos(x) + 1/tan²(x) = cos(x) + cot(x)

. . . . .[cos(x) + 1] / tan²

Or something else?

Is the right-hand side one of the following?

. . . . .cos(x) / sec(x) - 1 = cos²(x) - 1

. . . . .cos(x) / [sec(x) - 1]

Or something else?

What were your steps?

Please be complete. Thank you!

Eliz.

 

[posenkid88]

Left hand side is cosx+1/tan^2x and the right hand side is cosx/secx-1 and they equal each other but I cant find out what to do next. I take the left side and switch it to cosx+1/sec^2x-1 but i dont know where to go from there...like do i factor it out or what?

 

[stapel]

Left hand side is cosx+1/tan^2x and the right hand side is cosx/secx-1

Alright; so you have the following:

. . . . .\(\displaystyle \L \cos{(x)}\, +\, \frac{1}{\tan^2{(x)}}\, =\, \frac{\cos{(x)}}{\sec{(x)}}\, -\, 1\)

Since this is not actually an identity, you only need to find an x-value for which the two sides are unequal, and you're done!

Eliz.

 

[posenkid88]

I think i may of mislead the problem. On the left side it is (cosx+1)/tan^2x on the right side it is cosx/(secx-1).

Next i went (cosx+1)/(sec^2x-1) = cosx/(secx-1)

Then i factor and got (cosx+1)/(secx+1)(secx-1) = cosx/(secx-1)

But im lost at what to do next???

P.S. The x's = Theta

 

[Mrspi]

I think i may of mislead the problem. On the left side it is (cosx+1)/tan^2x on the right side it is cosx/(secx-1).

Next i went (cosx+1)/(sec^2x-1) = cosx/(secx-1)

Then i factor and got (cosx+1)/(secx+1)(secx-1) = cosx/(secx-1)

But im lost at what to do next???

P.S. The x's = Theta

Ok.....how about this....multiply both numerator and denominator of the right side by (sec x + 1).....

(cos x + 1) / (sec x + 1)(sec x - 1) = cos x(sec x + 1) / (sec x - 1)(sec x + 1)

Simplify the numerator on the right side: cos x * sec x + cos x * 1, or 1 + cos x

So...the right side is now (1 + cos x) / (sec x - 1)(sec x + 1)

Isn't that what you've got on the left side?? DONE.

 

[posenkid88]

Thanks that would work...but all i'm allowed to do it one side not both. So one side has to stay the same and i gotta match the other side with it.

 

[Mrspi]

Thanks that would work...but all i'm allowed to do it one side not both. So one side has to stay the same and i gotta match the other side with it.

Well, not exactly....

you can work with ONE side....and then you can work with the OTHER side independently, as long as you don't do anything that affects both sides....which we didn't here.

HOWEVER...if you insist.....

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

I'll work with the RIGHT SIDE ONLY......

What you see here is what I'm doing with the RIGHT SIDE:

cos x (sec x + 1) / [(sec x - 1)(sec x + 1)]

(cos x sec x + cos x) / (sec^2 x - 1)

(1 + cos x) / tan^2 x

DONE.....

 

[posenkid88]

Ah thank you very much.

 

[Mrspi]

an addendum....

You can work with one side of an identity for a while.

Then, you can work with the other side.

If you can get one side to look like the other side, that is an acceptable way to verify an identity.

You may want to check this with your teacher.