# Trig Identites

• Oct 18th 2010, 02:05 PM
RTC1996
Trig Identites
Prove: secθ - cosθ = sinθcosθ

My Work:

I know that secθ = 1/cosθ.

So, I replace sine theta with 1/cos theta.

(1/cosθ) - cosθ = sinθcosθ

On the left side, I simply apply the rules for subtraction of fractions.

The left side becomes (1 - cos^2θ)/cosθ.

The equation now looks like this:

(1 - cos^2θ)/cosθ = sinθcosθ

Where do I go from there?
• Oct 18th 2010, 02:24 PM
pickslides
$1-\cos^2\theta = \sin^2\theta$
• Oct 18th 2010, 07:26 PM
Educated
It seems to me that the trigonometric identity isn't true.

$\sec \theta - \cos \theta \ne \sin \theta \cos \theta$

Is the trigonometric identity supposed to be : $\sec \theta - \cos \theta = \sin \theta \tan \theta = \frac{\sin^2 x}{\cos x}$ ??

And also if you are proving trigonometric identities, you shouldn't have the answer at the end.
• Oct 20th 2010, 06:46 AM
Yudhishm
Here is some formulae:

$\tan A = \frac{\sin A}{\cos A}$

$\sec A = \frac{1}{\cos A}$

$cosec A = \frac{1}{\sin A}$

$\cot A = \frac{1}{\tan A} = \frac{\cos A}{\sin A}$

$\sin^2 \theta = 1 - \cos^a \theta$

$\tan^2 \theta = \sec^2 \theta - 1$

$\cot^2 \theta = cosec^2 \theta -1$

Hope this helped.... (Nerd)
• Oct 21st 2010, 08:19 AM
tempq1
Quote:

Originally Posted by RTC1996
Prove: secθ - cosθ = sinθcosθ

My Work:

I know that secθ = 1/cosθ.

So, I replace sine theta with 1/cos theta.

(1/cosθ) - cosθ = sinθcosθ

On the left side, I simply apply the rules for subtraction of fractions.

The left side becomes (1 - cos^2θ)/cosθ.

The equation now looks like this:

(1 - cos^2θ)/cosθ = sinθcosθ

Where do I go from there?

hello,
it seems that u cant solve the following identities as for proving, u have to make LHS=RHS
and u can test out by substituting a number using the calculator and if u try,the LHS =/= RHS.
To prove the equation,the LHS must be = to RHS.
correct me if im wrong.
• Oct 22nd 2010, 12:27 PM
RTC1996
Quote:

Originally Posted by Educated
It seems to me that the trigonometric identity isn't true.

$\sec \theta - \cos \theta \ne \sin \theta \cos \theta$

Is the trigonometric identity supposed to be : $\sec \theta - \cos \theta = \sin \theta \tan \theta = \frac{\sin^2 x}{\cos x}$ ??

And also if you are proving trigonometric identities, you shouldn't have the answer at the end.

I agree that this trig problem has no solution.
• Oct 22nd 2010, 12:28 PM
RTC1996
I don't think there is a solution in this case.
• Oct 22nd 2010, 12:36 PM
RTC1996
Quote:

Originally Posted by pickslides
$1-\cos^2\theta = \sin^2\theta$

I don't think there is a solution in this case.