1. ## 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?

2. $\displaystyle 1-\cos^2\theta = \sin^2\theta$

3. It seems to me that the trigonometric identity isn't true.

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

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

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

4. Here is some formulae:

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

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

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

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

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

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

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

Hope this helped....

5. 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.

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

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

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

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.

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

8. Originally Posted by pickslides
$\displaystyle 1-\cos^2\theta = \sin^2\theta$
I don't think there is a solution in this case.