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Trigonometric Ratios of Compound Angles

If a tanα +...

Question

If a $tan α + b tan β = (a + b) tan (\frac{α + β}{2})$ , where $α \neq β$ , prove that $a cos β = b cos α$ ,

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Solution

we have given,
$\begin{matrix} a tan α + b tan β = (a + b) tan (\frac{α + β}{2}) \Rightarrow tan α + b tan β = a tan (\frac{α + β}{2}) + b tan (\frac{α + β}{2}) \Rightarrow a [tan α - tan (\frac{α + β}{2})] = b [tan (\frac{α + β}{2}) - tan β] u s i n g t h e f o r m u l a : - -------------------- - tan A - tan B = \frac{sin (A - B)}{cos A . cos B} N o w, s u b s t i t u t i n g t h e f o r m u l a b o t h s i d e : \Rightarrow a \frac{sin (α - \frac{α + β}{2})}{cos α . cos (\frac{α + β}{2})} = b \frac{sin (\frac{α + β}{2} - β)}{cos β . cos (\frac{α + β}{2})} \Rightarrow a \frac{sin (α - \frac{α + β}{2})}{cos α} = b \frac{sin (\frac{α + β}{2} - β)}{cos β} \Rightarrow a \frac{sin (\frac{α - β}{2})}{cos α} = b \frac{sin (\frac{α - β}{2})}{cos β} \Rightarrow \frac{a}{cos α} = \frac{b}{cos β} ∴ a cos β = b cos α p r o v e . \end{matrix}$

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