# Establish the trig identity

• Oct 8th 2009, 06:20 PM
rebelspartan
Establish the trig identity
$\displaystyle \frac{\sec\theta - 1}{\tan^3\theta} = \frac{\cot\theta}{\sec\theta + 1}$

I need help verifying that these two expressions are equal (getting one side of the equation to look like the other without using properties of equality, just trig identities and simplifying one side).

Thanks.
• Oct 8th 2009, 08:31 PM
Hello rebelspartan

Welcome to Math Help Forum!
Quote:

Originally Posted by rebelspartan
$\displaystyle \frac{\sec\theta - 1}{\tan^3\theta} = \frac{\cot\theta}{\sec\theta + 1}$

I need help verifying that these two expressions are equal (getting one side of the equation to look like the other without using properties of equality, just trig identities and simplifying one side).

Thanks.

Three things you need here:

• $\displaystyle \sec^2\theta = 1+\tan^2\theta$ (1)

• $\displaystyle \cot\theta = \frac{1}{\tan\theta}$ (2)

• $\displaystyle a^2-b^2 = (a+b)(a-b)$ (3)

So:

1. Express the denominator (of the LHS) as $\displaystyle \tan\theta\times\tan^2\theta$, and then use (1) to express $\displaystyle \tan^2\theta$ in terms of $\displaystyle \sec^2\theta$
2. Use (3) to factorise the denominator
3. Cancel (simplify) the whole fraction
4. Use (2) on the remaining term in $\displaystyle \tan\theta$

... and you're there.

Can you do that?

• Oct 8th 2009, 08:42 PM
rebelspartan
Great, I got it! Thanks!
• Oct 8th 2009, 11:00 PM
pacman
http://www.mathhelpforum.com/math-he...c64ea8b5-1.gif

rebelspartan, this is less rigorous but, what if we cross multiply? let t = theta

--------------------------------------------------------------------------------------------------------------

(sec t - i)(sec t + 1) = cot t tan^3 t = (1/tan t)(tan t)(tan^2 t), that red thing cancels out.

the LHS is the difference of 2 squares, (sec t - 1)(sec t + 1) = sec^2 t - 1

LHS = RHS, indeed sec^2 t - 1 = tan^2 t,

remember this PYTHAGOREAN IDENTITY: tan^2 t + 1 = sec^2 t
• Oct 8th 2009, 11:27 PM
Hello everyone
Quote:

Originally Posted by pacman
http://www.mathhelpforum.com/math-he...c64ea8b5-1.gif

rebelspartan, this is less rigorous but, what if we cross multiply? let t = theta

--------------------------------------------------------------------------------------------------------------

(sec t - i)(sec t + 1) = cot t tan^3 t = (1/tan t)(tan t)(tan^2 t), that red thing cancels out.

the LHS is the difference of 2 squares, (sec t - 1)(sec t + 1) = sec^2 t - 1

LHS = RHS, indeed sec^2 t - 1 = tan^2 t,

remember this PYTHAGOREAN IDENTITY: tan^2 t + 1 = sec^2 t

Very bad style! Don't start off by assuming the thing you want to prove, and end up proving something that is known to be true. This is completely the wrong way round!

Having said that, if you can do this, then you should be able to 'reverse-engineer' your proof so that it all reads in the right direction and is therefore mathematically sound.

• Oct 8th 2009, 11:57 PM
pacman
i agree with you grandad, my above computation is really mathematically unsound, because i assumed it already as an identity, but i experimented with it by hindsight . . . . and it works.

http://www.mathhelpforum.com/math-he...c64ea8b5-1.gif,

formal Proof:

Consider the RHS, that is (cot t)/(sec t + 1) only

multiply and divide by (sec t - 1)

(cot t)/(sec t + 1) = [(cot t)/(sec t + 1)][(sec t - 1)/(sec t - 1)],

use the identity: 1 + tan^2 t = sec^2 t and simplify,

(cot t)/(sec t + 1) = (cot t)(sec t - 1)/(sec^2 t - 1)

(cot t)/(sec t + 1) = (cot t)(sec t - 1)/(tan^2 t)

(cot t)/(sec t + 1) = (cot t)(sec t - 1)/(tan^2 t)

(cot t)/(sec t + 1) = (1/tan t)(sec t - 1)/(tan^2 t)

(cot t)/(sec t + 1) = (sec t - 1)/[(tan t)(tan^2 t)]

(cot t)/(sec t + 1) = (sec t - 1)/(tan^3 t) = LHS.

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• Oct 9th 2009, 04:19 AM