Prove this trig identity.

**mwok** · November 26th 2008, 06:57 AM

Can someone solve it step by step. I tried two ways to solve it but didn't work.

**Smancer** · November 26th 2008, 07:04 AM

Its not clear to me what your question is, what are you supposed to show?

It is clear that your steps are not correct. Can you state the question exactly how it is given to you?

**skeeter** · November 26th 2008, 07:10 AM

$\frac{\sec^2{x}}{2 - \sec^2{x}} =$

$\frac{1 + \tan^2{x}}{2 - (1 + \tan^2{x})} =$

$\frac{1 + \tan^2{x}}{1 - \tan^2{x}}$

multiply by $\frac{\cos^2{x}}{\cos^2{x}}$ ...

$\frac{\cos^2{x} + \sin^2{x}}{\cos^2{x} - \sin^2{x}} =$

$\frac{1}{\cos(2x)} =$

$\sec(2x)$

**Soroban** · November 26th 2008, 07:15 AM

Hello, mwok!

Is this it? . $\sec2\theta \:=\:\frac{\sec^2\!\theta}{2-\sec^2\!\theta}$

The left side is: . $\sec2\theta \;=\;\frac{1}{\cos2\theta} \;=\;\frac{1}{2\cos^2\!\theta - 1}$

Divide top and bottom by $\cos^2\!\theta\!:\quad \frac{\dfrac{1}{\cos^2\!\theta}}{\dfrac{2\cos^2\!\ theta}{\cos^2\!\theta} - \dfrac{1}{\cos^2\!\theta}} \;=\;\frac{\sec^2\!\theta}{2 - \sec^2\!\theta}$

**mwok** · November 26th 2008, 08:15 AM

Thanks!

**mwok** · November 26th 2008, 09:10 AM

Originally Posted by skeeter

$\frac{\sec^2{x}}{2 - \sec^2{x}} =$

$\frac{1 + \tan^2{x}}{2 - (1 + \tan^2{x})} =$

$\frac{1 + \tan^2{x}}{1 - \tan^2{x}}$

multiply by $\frac{\cos^2{x}}{\cos^2{x}}$ ...

$\frac{\cos^2{x} + \sin^2{x}}{\cos^2{x} - \sin^2{x}} =$

$\frac{1}{\cos(2x)} =$

$\sec(2x)$

Why do you multiply with the cos/sin?

**skeeter** · November 26th 2008, 09:38 AM

Originally Posted by mwok

Why do you multiply with the cos/sin?

I didn't ... I multiplied by $\frac{\cos^2{x}}{\cos^2{x}}$

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