Trigo identity

**elitewarr** · February 6th 2009, 12:00 AM

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.

**skeeter** · February 6th 2009, 04:24 PM

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 ...

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

**elitewarr** · February 6th 2009, 10:10 PM

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

**running-gag** · February 7th 2009, 01:29 AM

Hi

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

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

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

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

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

**elitewarr** · February 7th 2009, 05:13 AM

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

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