I run into a problem when I'm trying to prove how $\tan^2x+1 = \sec^2x$, and $1+\cot^2x=\csc^2x$

I understand that $\sin^2x+\cos^2x = 1$. (To my understanding 1 is the Hypotenuse, please correct me if I'm wrong). If referring to a Pythagorean triangle, let's say a triangle where $a=3$, $b=4$, and $c=5$, or $a=\cos$ $b=\sin$ and $c=\text{hypotenuse}$.

$3^2 + 4^2 = 5^2$. Which is true and proves that this identity work. To my understanding, that is how this identity work.

However, when I try to make sense of the $\tan^2x+1 = \sec^2x$ and $1+\cot^2x=\csc^2x$ identity, using the triangle example from above, it doesn't work. For example, here's how I did it on paper,

$\tan^2x + 1 = {\sin^2x\over \cos^2x} + 1$, so I would get using the triangle above, ${4^2\over 3^2} + 1 = {16\over 9} + 1 = 2.777777778$

Now on the right hand side, $\sec^2x, = {1\over \cos^2x} = {1\over 3^2} = .1111111111$. The answer I get from $\tan^2x+1$ DOES NOT EQUAL the answer I get from $\sec^2x$. I think I may be misunderstanding a critical part here that I can't really pinpoint.

However, I do know that if I prove $\tan^2x+1 = \sec^2x$ using just the identity itself it does work, for example,

$\tan^2x+1 = {\sin^2x\over \cos^2x} + 1 = (\text{after some simplification}) = \sin^2x + {\cos^2x \over \cos^2x} = {1\over \cos^2x} = \sec^2x$.

The same issue happens with $1+\cot^2x=\csc^2x$.

To my understanding, $\sin^2+\cos^2=1$ is the same as $a^2+b^2=c^2$. Am I right or wrong? I think there is a huge concept that I am missing between the UNIT CIRCLE and just TRIANGLES.

  • 1
    $\begingroup$ cos(x)=3/5 and sin(x)=4/5; I don't think you are using these definitions correctly. cos=adj/hyp and sin=opp/hyp $\endgroup$
    – randomgirl
    Feb 26 '15 at 1:31
  • $\begingroup$ Try reviewing your statement " let's say a triangle where a=3, b=4, and c=5, or a=cos b=sin and c=hypotenuse.". $\endgroup$
    – John Joy
    Feb 27 '15 at 21:37
  • $\begingroup$ If any of the answeres below were useful to you, then you should upvote all answers you find useful and accept the one that was most useful to you. It is a way to show that you have found the answer to your question and it shows your appreciation. Now it seems like you still need help. If answers are not useful to you, then it helps if you say why not. This helps others to help you. For more information read this. $\endgroup$
    – gebruiker
    Mar 15 '16 at 17:52
  • $\begingroup$ @gebruiker: FWIW, OP can accept the best answer, but at $6$ points, does not yet have enough reputation to upvote the others. $\endgroup$
    – Brian Tung
    Mar 15 '16 at 17:58

Your issue is your understanding of the relationship between your basic trigonometry values and the sides of a triangle. You must keep in mind that the values of trigonometric functions are utilising the sides of your triangles in the following manner: $\sin(x) = \frac{opp}{hyp}$, $\cos(x) = \frac{adj}{hyp}$, and $\tan(x) = \frac{\sin(x)}{\cos(x)}$.

Let us use your example of a $3$, $4$, $5$ triangle to prove our identities.

triangle, much wow Let us go ahead and figure out our values of $\cos(x)$, $\sin(x)$, and $\tan(x)$:

$$\sin(x) = \frac{4}{5}\\ \cos(x) = \frac{3}{5}\\ \tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}.$$

Now that we have figured out our values, we may plug them into our identities:

$$\sin^2(x) + \cos^2(x) = 1\\ \left(\frac{4}{5}\right)^2 + \left(\frac{3}{5}\right)^2 = 1\\ \frac{16}{25} + \frac{9}{25} = 1\\ \frac{25}{25} = 1$$


$$1 + \cot^2(x) = \csc^2(x)\\ 1 + \left(\frac{1}{\frac{4}{3}}\right)^2 = \left(\frac{1}{\frac{4}{5}}\right)^2\\ 1 + \left(\frac{3}{4}\right)^2 = \left(\frac{5}{4}\right)^2\\ 1 + \frac{9}{16} = \frac{25}{16}\\ \frac{16}{16} + \frac{9}{16} = \frac{25}{16}\\ \frac{25}{16} = \frac{25}{16}$$

Attempt to prove $\tan^2(x) + 1 = \sec^2(x)$ by yourself to see if you fully understand.

A quick re-cap on the relationship of the three identities you proposed. Assuming you understand where $\sin^2(x) + \cos^2(x) = 1$ originated from, you should have no problems understanding where the next two identities originated from.

$1 + \cot^2(x) = \csc^2(x)$:

$$\sin^2(x) + \cos^2(x) = 1\\ \frac{\sin^2(x)}{\sin^2(x)} + \frac{\cos^2(x)}{\sin^2(x)} = \frac{1}{\sin^2(x)}\\ 1 + \cot^2(x) = \csc^2(x)$$

$\tan^2(x) + 1 = \sec^2(x)$:

$$\sin^2(x) + \cos^2(x) = 1\\ \frac{\sin^2(x)}{\cos^2(x)} + \frac{\cos^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)}\\ \tan^2(x) + 1 = \sec^2(x)$$


The first identity works because $\frac{3}{5}^2+\frac{4}{5}^2=\frac{5}{5}^2$. When thinking of these trig ratios, you have to think of them in terms of both triangle sides you are interested in. Not merely one leg being equal to $\sin$, for example.

The beauty of the "unit" circle is that with the hypotenuse being $1$, in this case, we can think of the legs of the triangle being the actual value of $\sin$ and $\cos$, because the ratio of that length to $1$ is just that length.

You can workout the other identities if you express tan as $\frac{4}{3}$, $\csc$ as $\frac{5}{4}$, etc.


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