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

Don't have a clue what to do.

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- Mar 4th 2010, 08:53 AM #1

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- Mar 4th 2010, 09:11 AM #2

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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

- Mar 4th 2010, 09:45 AM #3

- Mar 4th 2010, 09:59 AM #4

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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!

- Mar 4th 2010, 06:18 PM #5

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- Mar 4th 2010, 06:21 PM #6

- Mar 4th 2010, 06:28 PM #7

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- Mar 4th 2010, 06:37 PM #8

- Mar 4th 2010, 06:45 PM #9

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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.

- Mar 4th 2010, 06:49 PM #10
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.

- Mar 4th 2010, 06:50 PM #11

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- Mar 4th 2010, 06:51 PM #12

- Mar 4th 2010, 06:52 PM #13

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- Mar 4th 2010, 06:53 PM #14

- Mar 4th 2010, 07:39 PM #15

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