1. ## trig problems:

1. prove that $\frac{ \cos A +1-\sin A }{ \cos A - 1+\sin A} = \frac{1+\cos A}{\sin A}$

2.if $\tan A=n \tan B$ ,then find the maximum value of $\tan^2 (A-B)$

2. what have you done? show your work.

3. yup!got the first one...but have no idea on the second one.

(1) $tan(A-B)=\frac{tanA-tanB}{1+tanAtanB}$.

Now let we divide (1) by $tanB$:

(2) $\frac{tan(A-B)}{tanB}=\frac{\frac{tanA}{tanB}-1}{1+tanAtanB}$

We given that: (3) $\frac{tanA}{tanB}=n$

Putting (3) in (2):

$\frac{tan(A-B)}{tanB}=\frac{n-1}{\frac{1}{tanB}+ntanB}$

Now I will assume that $tanB>0$, and will sign $tanB=t$.

Let us now look at: $\frac{1}{tanB}+ntanB=\frac{1}{t}+nt$

Using: Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia

$\frac{\frac{1}{t}+nt}{2}\geq \sqrt{n}$

Or:

${\frac{1}{t}+nt\geq 2\sqrt{n}$

Hence $min\left \{ \frac{1}{tanB}+ntanB \right \}= 2\sqrt{n}$

Maximum value of $tan(A-B)=\frac{n-1}{ 2\sqrt{n}}$

Therefor $max\left \{tan^2(A-B) \right \}=\frac{(n-1)^2}{4n}$.

5. I would like to know how you went from this:

$\dfrac{tan(A-B)}{tanB}=\dfrac{\frac{tanA}{tanB}-1}{1+tanAtanB}$

to this:

$\dfrac{tan(A-B)}{tanB}=\dfrac{n-1}{\frac{1}{tanB}+ntanB}$

Concerning the denominator...

Shouldn't it be:

$\dfrac{tan(A-B)}{tanB}=\dfrac{n-1}{1+ntan^2B}$

Or

$\dfrac{tan(A-B)}{tanB}=\dfrac{\frac{n-1}{tanB}}{\frac{1}{tanB}+ntanB}$

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# If tanA=ntanB find maximum value of square of tan(A-B)

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