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 '14 at 15:06
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Please let us know your progress so far. Hint: Expand both Nr and Dr – Sandeep Thilakan Feb 10 '14 at 15:07
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Related : http://math.stackexchange.com/questions/668619/proving-a-trig-identity – lab bhattacharjee Feb 10 '14 at 16:37
3 Answers
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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 $(\cos a\cos b)...$
amWhy
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We start with $\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}$:
$\frac{1+tan(a)tan(b)}{1-tan(a)tan(b)}=\frac{1+\frac{sin (a)sin (b)}{cos (a)cos (b)}}{1-\frac{sin (a)sin (b)}{cos (a)cos (b)}}=\frac{\frac{cos (a)cos (b)+sin (a)sin (b)}{cos (a)cos (b)}}{\frac{cos (a)cos (b)-sin (a)sin (b)}{cos (a)cos (b)}}=\frac{cos (a)cos (b)+sin (a)sin (b)}{cos (a)cos (b)-sin (a)sin (b)}=\frac{cos(a-b)}{cos(a+b)}$
Bobby
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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*}
BlackAdder
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