1. Proving identities

a) Prove the Identity cot 1/2 A - tan 1/2 A = 2cotA

b) By choosing a suitable numerical value for A, show that tan15 is a root of the quadratic equations t^2 + 2 sqrt.3t -1 = 0

thanks.

2. Let B = A/2

$\displaystyle cot B - tan B = \frac { cos B}{sin B } -\frac {sin B}{cos B}$

$\displaystyle = \frac { cos^2 B - sin^2B}{sin B cos B}$

$\displaystyle = \frac { cos2 B}{(sin 2B)/2}$

$\displaystyle = 2 cot 2B$

3. Originally Posted by Oasis1993
a) Prove the Identity cot 1/2 A - tan 1/2 A = 2cotA
Originally Posted by Oasis1993
thanks.

Hi all

4. Originally Posted by qmech
Let B = A/2

$\displaystyle cot B - tan B = \frac { cos B}{sin B } -\frac {sin B}{cos B}$

$\displaystyle = \frac { cos^2 B - sin^2B}{sin B cos B}$

$\displaystyle = \frac { cos2 B}{(sin 2B)/2}$

$\displaystyle = 2 cot 2B$
sorry sir
i didnt see your solution
mrmohamed

5. Originally Posted by Oasis1993
a) Prove the Identity cot 1/2 A - tan 1/2 A = 2cotA

b) By choosing a suitable numerical value for A, show that tan15 is a root of the quadratic equations t^2 + 2 sqrt.3t -1 = 0

thanks.
HI

$\displaystyle \cot \frac{1}{2}A-\tan \frac{1}{2}A=\frac{\cos \frac{1}{2}A}{\sin \frac{1}{2}A}-\frac{\sin \frac{1}{2}A}{\cos \frac{1}{2}A}$

$\displaystyle \frac{\cos^2 \frac{1}{2}A-\sin^2 \frac{1}{2}A}{\sin \frac{1}{2}A\cos \frac{1}{2}A}$

$\displaystyle \frac{\cos A}{\sin \frac{1}{2}A\cos \frac{1}{2}A}\cdot \frac{2}{2}$

$\displaystyle \frac{2\cos A}{\sin A}$

$\displaystyle 2\cos A$ (proved)

Now , from this identity ..

$\displaystyle \tan \frac{1}{2}A=\cot \frac{1}{2}A-2\cot A$

$\displaystyle \tan \frac{1}{2}(30)=\cot \frac{1}{2}(30)-2\cot 30$

$\displaystyle \tan 15=\frac{1}{\tan 15}-\frac{2}{\tan 30}$

$\displaystyle \tan 15=\frac{1}{\tan 15}-2\sqrt{3}$

$\displaystyle \tan^2 15+2\sqrt{3}\tan 15-1=0$

let $\displaystyle \tan 15$ be t

$\displaystyle t^2+2\sqrt{3}t-1=0$

Then by using the quadratic formula

$\displaystyle t=\frac{-2\sqrt{3}\pm \sqrt{16}}{2}$

$\displaystyle t=-\sqrt{3}\pm 2$

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tan 1/2a cot 1/2a

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