Prove(2)

**Punch** · January 21st 2010, 11:06 PM

Prove $ \frac{1+sec2A}{tan2A}=cotA $

**Defunkt** · January 22nd 2010, 03:47 AM

Originally Posted by Punch

Prove $ \frac{1+sec2A}{tan2A}=cotA $

$\frac{1+sec2\alpha}{tan2\alpha} = \frac{1 + \frac{1}{cos(2\alpha)}}{\frac{sin(2\alpha)}{cos(2\ alpha)}} = \frac{cos(2\alpha)+1}{cos(2\alpha)} \cdot \frac{cos(2\alpha)}{sin(2\alpha)} =$

$= \frac{cos(2\alpha)+1}{sin(2\alpha)} = \frac{cos^2(\alpha) - sin^2(\alpha) + cos^2(\alpha)+sin^2(\alpha)}{2sin(\alpha)cos(\alph a)} = \frac{2cos^2(\alpha)}{2sin(\alpha)cos(\alpha)} =$ $\frac{cos(\alpha)}{sin(\alpha)} = cot(\alpha)$

Identities used were:

$(1)~ cos(2\alpha) = cos^2(\alpha)-sin^2(\alpha)$
$(2) ~sin(2\alpha) = 2sin(\alpha)cos(\alpha)$
$(3) ~1 = sin^2(\alpha)+cos^2(\alpha)$

These really are not difficult... simply expand and use a list of trigonometric identities

**Punch** · January 22nd 2010, 03:52 AM

Originally Posted by Defunkt

$\frac{1+sec2\alpha}{tan2\alpha} = \frac{1 + \frac{1}{cos(2\alpha)}}{\frac{sin(2\alpha)}{cos(2\ alpha)}} = \frac{cos(2\alpha)+1}{cos(2\alpha)} \cdot \frac{cos(2\alpha)}{sin(2\alpha)} =$

$= \frac{cos(2\alpha)+1}{sin(2\alpha)} = \frac{cos^2(\alpha) - sin^2(\alpha) + cos^2(\alpha)+sin^2(\alpha)}{2sin(\alpha)cos(\alph a)} = \frac{2cos^2(\alpha)}{2sin(\alpha)cos(\alpha)} =$ $\frac{cos(\alpha)}{sin(\alpha)} = cot(\alpha)$

Identities used were:

$(1)~ cos(2\alpha) = cos^2(\alpha)-sin^2(\alpha)$
$(2) ~sin(2\alpha) = 2sin(\alpha)cos(\alpha)$
$(3) ~1 = sin^2(\alpha)+cos^2(\alpha)$

These really are not difficult... simply expand and use a list of trigonometric identities

Sometimes, I am really stucked at proving, I have a phobia of it and was stucked when I got to \frac{cos2A-1}{sin2A}... Seems like I need more practise in proving, give me tips/comments if you have any

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