# Thread: Trigo identity

1. ## Trigo identity

Hello, I need help with this problem
$\frac{\sec{8\theta}-1}{\sec{4\theta}-1} = \frac{\tan{8\theta}}{\tan{2\theta}}$

Thank you.

2. Originally Posted by elitewarr
Hello, I need help with this problem

$\frac{\sec{8\theta}-1}{\sec{4\theta}-1} = \frac{\tan{8\theta}}{\tan{4\theta}}$
as posted, the equation is not an identity ...

this would be ...

$\frac{\sec^2(8\theta)-1}{\sec^2(4\theta)-1} = \frac{\tan^2(8\theta)}{\tan^2(4\theta)}$

3. My mistake, the tan4 theta should be tan 2 theta. I edited the post.

4. Hi

$\frac{\sec{8\theta}-1}{\sec{4\theta}-1} \:\frac{\tan{2\theta}}{\tan{8\theta}} = \frac{\cos{4\theta}}{\cos{8\theta}} \frac{1-\cos{8\theta}}{1-\cos{4\theta}} \:\frac{\tan{2\theta}}{\tan{8\theta}}$

$\frac{\sec{8\theta}-1}{\sec{4\theta}-1} \:\frac{\tan{2\theta}}{\tan{8\theta}} = \frac{\cos{4\theta}}{\cos{8\theta}} \frac{2 \sin^2{4\theta}}{2 \sin^2{2\theta}} \:\frac{\sin{2\theta}}{\sin{8\theta}}\:\frac{\cos{ 8\theta}}{\cos{2\theta}}$

$\frac{\sec{8\theta}-1}{\sec{4\theta}-1} \:\frac{\tan{2\theta}}{\tan{8\theta}} = \frac{\cos{4\theta} \sin^2{4\theta}}{\sin{2\theta} \sin{8\theta} \cos{2\theta}}$

$\frac{\sec{8\theta}-1}{\sec{4\theta}-1} \:\frac{\tan{2\theta}}{\tan{8\theta}} = \frac{\cos{4\theta} \sin^2{4\theta}}{\sin{2\theta}\: 2 \sin{4\theta} \cos{4\theta} \cos{2\theta}}$

$\frac{\sec{8\theta}-1}{\sec{4\theta}-1} \:\frac{\tan{2\theta}}{\tan{8\theta}} = \frac{\sin{4\theta}}{2 \sin{2\theta} \cos{2\theta}} = 1$

5. Wow..Thanks..I thought for more than 1 hour and I still could not find the solution.