# Having trouble verifying a trig identity.

## Homework Statement

$\frac{cos^{2}t+tan^{2}t -1}{sin^{2}t} = tan^{2}t$

## Homework Equations

Here are all the trig identities we know up to this point (the one's that we have learned so far, obviously we derive many others from these when verifying identities).

Pythagorean Identities:
$sin^{2}x + cos^{2}x = 1$
$tan^{2}x + 1 = sec^{2}x$
$cot^{2}x + 1 = csc^{2}x$

Reciprocal Identities:
$\frac{sinx}{cosx} = tanx$
$\frac{1}{sinx} = cosx$

Even-Odd Identities:
$sin(-x) = -sinx$
$cos(-x) = cosx$
$tan(-x) = -tanx$

## The Attempt at a Solution

I've tried this a couple different ways, but still can't figure it out. I'll only show the 3 attempts that I feel came the closest, as these are rather time consuming to type up (But I do think this site does math typing the best I've seen so far). By the third attempt, I began to wonder if this is even a possible identity.

Also, zooming in on your browser will make some of the fractions more readable. Most browsers use the shortcut "Ctrl +" for this.

The first attempt uses the left side (the side with the fraction)
$\frac{cos^{2}t}{sin^{2}t} + \frac{tan^{2}t}{sin^{2}t} - \frac{1}{sin^{2}t}$

$cot^{2}t + \frac{tan^{2}t}{1-cos^{2}t} - csc^{2}t$

$csc^{2}t - 1 - csc^{2}t + \frac{tan^{2}t}{1 - cos^{2}x}$

$\frac{tan^{2}t}{1 - \frac{sin^{2}t}{tan^{2}t}} -1$

$\frac{tan^{2}t}{\frac{tan^{2}t}{tan^{2}t} -\frac{sin^{2}t}{tan^{2}t}} - 1$

$\frac{tan^{4}t}{tan^{2}t - sin^{2}t} - 1$

$\frac{tan^{4}t - tan^{2}t + sin^{2}t}{tan^{2}t - sin^{2}t}$
...and this is where I give up with this method, not seeing how this will = $tan^{2}t$

My second attempt I try the right side, that only has the tan
$sec^{2}t - 1$

$\frac {1}{cos^{2}t} - sin^{2}t - cos^{2}t$

$\frac{1}{cos^{2}t} - \frac{(cos^{2}t)(sin^{2}t)}{cos^{2}t} - \frac{cos^{4}t}{cos^{2}t}$

$\frac{1 - (cos^{2}t)(sin^{2}t) - cos^{4}t}{cos^{2}t}$

$\frac{-1(cos^{4}t + (cos^{2}t)(sin^{2}t) - cos^{2}t - sin^{2}t}{cos^{2}t}$

$\frac{-1(cos^{4}t + (cos^{2}t)(sin^{2}t) - cos^{2}t - (tan^{2}t)(cos^{2}t)}{cos^{2}t}$

$\frac{-1(cos^{2}t)(cos^{2}t + sin^{2}t - 1 - tan^{2}t)}{cos^{2}t}$

$-cos^{2}t - sin^{2}t + 1 + tan^{2}t)$
...and this is when I give up. As you can see I was grasping at straws toward the end trying to get it figured out.

Attempt number 3, this one makes me think that this isn't even an identity. Started with the fraction side again

$\frac{cos^{2}t + tan^{2}t - sin^{2}t - cos^{2}t}{sin^{2}t}$

$\frac{tan^{2}t - sin^{2}t}{sin^{2}t}$

$\frac{tan^{2}t}{sin^{2}t} - \frac{sin^{2}t}{sin^{2}t}$

$\frac{tan^{2}t}{(tan^{2}t)(\frac{1}{csc^{2}t})} - 1$

$\frac{tan^{2}t}{\frac{tan^{2}t}{csc^{2}t})} - 1$

$(tan^{2}t)(\frac{csc^{2}t}{tan^{2}t}) - 1$

$csc^{2}t - 1$

$cot^{2}t$
...And here is where I began to wonder if this is a possible identity.

Anyways, as you can hopefully see, I have tried this many different ways and still can't figure it out. I'm probably missing something simple, but I just don't see it. Any help is greatly appreciated.

Related Precalculus Mathematics Homework Help News on Phys.org
SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

$\frac{cos^{2}t+tan^{2}t -1}{sin^{2}t} = tan^{2}t$

## Homework Equations

Here are all the trig identities we know up to this point (the one's that we have learned so far, obviously we derive many others from these when verifying identities).

Pythagorean Identities:
$sin^{2}x + cos^{2}x = 1$
$tan^{2}x + 1 = sec^{2}x$
$cot^{2}x + 1 = csc^{2}x$

Reciprocal Identities:
$\frac{sinx}{cosx} = tanx$
$\frac{1}{sinx} = cosx$

Even-Odd Identities:
$sin(-x) = -sinx$
$cos(-x) = cosx$
$tan(-x) = -tanx$

## The Attempt at a Solution

I've tried this a couple different ways, but still can't figure it out. I'll only show the 3 attempts that I feel came the closest, as these are rather time consuming to type up (But I do think this site does math typing the best I've seen so far). By the third attempt, I began to wonder if this is even a possible identity.

Also, zooming in on your browser will make some of the fractions more readable. Most browsers use the shortcut "Ctrl +" for this.

The first attempt uses the left side (the side with the fraction)
$\frac{cos^{2}t}{sin^{2}t} + \frac{tan^{2}t}{sin^{2}t} - \frac{1}{sin^{2}t}$

$cot^{2}t + \frac{tan^{2}t}{1-cos^{2}t} - csc^{2}t$

$csc^{2}t - 1 - csc^{2}t + \frac{tan^{2}t}{1 - cos^{2}x}$
...
Hello kieth89. Welcome to PF !

Rather than changing sin2(t) to 1-cos2(t),

change tan2(t) to $\displaystyle \frac{\sin^2(t)}{\cos^2(t)}$

Then you're almost there !

Hello kieth89. Welcome to PF !

Rather than changing sin2(t) to 1-cos2(t),

change tan2(t) to $\displaystyle \frac{\sin^2(t)}{\cos^2(t)}$

Then you're almost there !
Aha! I see it now. Thank you very much, that problem really had me stuck.

Also, why was I able, on attempt 3, to get $cot^{2}t = tan^{2}t$? I'm new to trig stuff, and when I see something like that I think right away not possible (like 3 = 5 and stuff). Of course, then I check it by plugging the initial equations into the calculator and see that they do (approximately) equal each other. But why can those two be equal? I know that cot is just the tangent ratio flipped...I don't know, it just seems weird when I'm able to get that. I very well could have made an error in my calculations, and will look them over after I post this. Just seems weird.

SammyS
Staff Emeritus
Homework Helper
Gold Member
Aha! I see it now. Thank you very much, that problem really had me stuck.

Also, why was I able, on attempt 3, to get $cot^{2}t = tan^{2}t$? I'm new to trig stuff, and when I see something like that I think right away not possible (like 3 = 5 and stuff). Of course, then I check it by plugging the initial equations into the calculator and see that they do (approximately) equal each other. But why can those two be equal? I know that cot is just the tangent ratio flipped...I don't know, it just seems weird when I'm able to get that. I very well could have made an error in my calculations, and will look them over after I post this. Just seems weird.
$\displaystyle\sin^2(t)\ne \frac{tan^2(t)}{\csc^2(t)}$

$\displaystyle\sin^2(t)=tan^2(t)\cdot\cos^2(t) =\frac{tan^2(t)}{\sec^2(t)}$

$\displaystyle\sin^2(t)\ne \frac{tan^2(t)}{\csc^2(t)}$

$\displaystyle\sin^2(t)=tan^2(t)\cdot\cos^2(t) =\frac{tan^2(t)}{\sec^2(t)}$
Thought it would be something simple like that. Got too careless after the first couple tries. Anyways, thank you for all the help.