Verify tan^2(x/2) = secx-1/secx+1

Question

tkhunny · Answer

$\tan^{2}(x/2) = \sec(x) - \dfrac{1}{\sec(x)} + 1$

Are you SURE that's what you meant?  That IS what you wrote.

anonymous · Answer

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

tkhunny · Answer

I see.  Please remember your Order of Operations.  Use parentheses to clarify.

Hmmm...

$\cos(2x) = 2\cos^{2}(x) - 1 = 1 - 2\sin^{2}(x)$

Transforming...

$\cos(x) = 2\cos^{2}(x/2) - 1 = 1 - 2\sin^{2}(x/2)$

Solving for the squared half angles:

$\cos^{2}(x/2) = \dfrac{\cos(x)+1}{2}$

$\sin^{2}(x/2) = \dfrac{\cos(x)-1}{-2} = \dfrac{1 - \cos(x)}{2}$

Did that get us anywhere?

anonymous · Answer

Not really I just completely don't understand this stuff at all have been trying for 3 hours and am about ready to give up

tkhunny · Answer

You're breaking my heart, here.  Do you know this one:  $\tan(x) = \dfrac{\sin(x)}{\cos(x)}$?