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Someone can help me showing this identity?

$\frac{cos(a-b)}{cos(a+b)}=\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}$

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Start by using the angle-sum (difference) formulas for $\cos$. –  amWhy Feb 10 at 15:06
    
Please let us know your progress so far. Hint: Expand both Nr and Dr –  Sandeep Thilakan Feb 10 at 15:07
    

3 Answers 3

up vote 5 down vote accepted

Using the angle sum formula for $\cos$:

$$\frac{\cos(a-b)}{\cos(a+b)}= \frac{\cos a \cos b + \sin a \sin b}{\cos a \cos b - \sin a \sin b}$$

Hint: Now all you need to do is to divide the numerator and denominator by $(\cos a\cos b)...$

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Nice hint! and have a nice day! –  Sami Ben Romdhane Feb 11 at 15:29

We start with $\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}$:

$\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}=\frac{1+\frac{sin (a)sin (b)}{cos (a)cos (b)}}{1-\frac{sin (a)sin (b)}{cos (a)cos (b)}}=\frac{\frac{cos (a)cos (b)+sin (a)sin (b)}{cos (a)cos (b)}}{\frac{cos (a)cos (b)-sin (a)sin (b)}{cos (a)cos (b)}}=\frac{cos (a)cos (b)+sin (a)sin (b)}{cos (a)cos (b)-sin (a)sin (b)}=\frac{cos(a-b)}{cos(a+b)}$

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Hint:

\begin{align*} \frac{\cos(a-b)}{\cos(a+b)}&=\frac{\cos a\cos b+ \sin a\sin b}{\cos a\cos b- \sin a\sin b}\\ &=\frac{\displaystyle\frac{\cos a\cos b+ \sin a\sin b}{\cos a\cos b}}{\displaystyle\frac{\cos a\cos b- \sin a\sin b}{\cos a\cos b}}\\ &=\; ? \end{align*}

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