Show that 1+tan^2x = (cosx)^-2 where defined.

Don't have a clue what to do.

2. Originally Posted by jumba
Show that 1+tan^2x = (cosx)^-2 where defined.

Don't have a clue what to do.
To prove whether or not

$\displaystyle 1+tan^2x=\frac{1}{cos^2x}$

then... either express tanx in terms of cosx, or vice versa

Simplest is to express tan in terms of cos

$\displaystyle 1+\frac{sin^2x}{cos^2x}=\frac{cos^2x}{cos^2x}+\fra c{sin^2x}{cos^2x}$

A very important trigonometric identity completes the proof after you express that as a single fraction

3. Don't make it complicated -_-
$\displaystyle 1+tan^2(x)=sec^2(x)=\frac{1}{cos^2(x)}$

4. Excellent response, Miss !!

However, $\displaystyle \frac{1}{cosx}$ is "called" $\displaystyle sec(x)$

so we must prove that

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

without swopping between the alternative descriptions of the same trigonometric ratio.

You have highlighted a situation where marks could be easily lost in an exam!

5. Originally Posted by Miss
Don't make it complicated -_-
$\displaystyle 1+tan^2(x)=sec^2(x)=\frac{1}{cos^2(x)}$
I don't know if you understood, jumba,

the above is not a proof,
it's only a statement.

Whether you choose to call $\displaystyle \frac{1}{cosx}$ as $\displaystyle sec(x)$ or not, you still need to prove that it is $\displaystyle 1+tan^2(x)$

6. Originally Posted by Archie Meade
I don't know if you understood, jumba,

the above is not a proof,
it's only a statement.

Whether you choose to call $\displaystyle \frac{1}{cosx}$ as $\displaystyle sec(x)$ or not, you still need to prove that it is $\displaystyle 1+tan^2(x)$
Its an acceptable proof, not a statement !
Its a standard identity.
If you think like this, then you need to prove that $\displaystyle tan^2(x)=\frac{sin^2(x)}{cos^2(x)}$.

7. $\displaystyle sec(x)=\frac{1}{cos(x)}$

is a definition.

Please don't try to confuse the student.

8. Originally Posted by Archie Meade
$\displaystyle sec(x)=\frac{1}{cos(x)}$

is a definition.

Please don't try to confuse the student.

I did not confuse him
But what I wrote is an accpetable proof.
Am sure

9. It's not a proof.

$\displaystyle sec(x)$ is the name given to $\displaystyle \frac{1}{cos(x)}$

The question may be written

Prove $\displaystyle 1+tan^2(x)=\frac{1}{cos^2(x)}$

Or..

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

It's the same question!

Interposing the alternative name does not outline a proof.

10. Originally Posted by Archie Meade
It's not a proof.

$\displaystyle sec(x)$ is the name given to $\displaystyle \frac{1}{cos(x)}$

The question may be written

Prove $\displaystyle 1+tan^2(x)=\frac{1}{cos(x)}$

Or..

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

It's the same question!

Interposing the alternative name does not outline a proof.
Well, then $\displaystyle \frac{sin^2(x)}{cos^2(x)}$ is a name given to $\displaystyle tan^2(x)$. So prove that $\displaystyle tan^2(x)=\frac{sin^2(x)}{cos^2(x)}$.
and $\displaystyle \frac{cos(x)}{cos(x)}$ is a name given to 1.
....etc etc
You are moving around a circle.

11. Are you serious????

12. Originally Posted by Archie Meade
Are you serious????
Sure.

13. I reckon it's a case for a moderator then

14. Originally Posted by Archie Meade
I reckon it's a case for a moderator then
OK. Do it.
Question: Do you know how many proofs have trigonometric identities in Earth?
According to your posts: they are not proofs, they are just statements!

15. How do you show that

$\displaystyle sec(x)=\frac{1}{cos(x)}$

without only stating that

$\displaystyle sec(x)=\frac{1}{cos(x)}$ ?

or without just calling it an identity.

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