If a tan alpha + b tan beta = (a + b) tan lef | vedclass

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(D) Given: $a 	an \alpha + b 	an \beta = (a + b) 	an \left( \frac{\alpha + \beta}{2} ight)$ Rearranging the terms: $a \left( 	an \alpha - 	an \

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$\frac{b}{a}$

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(D) Given: $a 	an \alpha + b 	an \beta = (a + b) 	an \left( \frac{\alpha + \beta}{2} ight)$ Rearranging the terms: $a \left( 	an \alpha - 	an \left( \frac{\alpha + \beta}{2} ight) ight) = b \left( 	an \left( \frac{\alpha + \beta}{2} ight) - 	an \beta ight)$ Using the identity $	an A - 	an B = \frac{\sin(A - B)}{\cos A \cos B}$: $\frac{a \sin \left( \alpha - \frac{\alpha + \beta}{2} ight)}{\cos \alpha \cos \left( \frac{\alpha + \beta}{2} ight)} = \frac{b \sin \left( \frac{\alpha + \beta}{2} - \beta ight)}{\cos \left( \frac{\alpha + \beta}{2} ight) \cos \beta}$ $\frac{a \sin \left( \frac{\alpha - \beta}{2} ight)}{\cos \alpha} = \frac{b \sin \left( \frac{\alpha - \beta}{2} ight)}{\cos \beta}$ Since $\alpha - \beta 
eq 2n\pi$,$\sin \left( \frac{\alpha - \beta}{2} ight) 
eq 0$. Dividing both sides by $\sin \left( \frac{\alpha - \beta}{2} ight)$: $\frac{a}{\cos \alpha} = \frac{b}{\cos \beta}$ Therefore,$\frac{\cos \beta}{\cos \alpha} = \frac{b}{a}$.

If $a \tan \alpha + b \tan \beta = (a + b) \tan \left( \frac{\alpha + \beta}{2} \right)$ and $\alpha - \beta \neq 2n\pi$,then $\frac{\cos \beta}{\cos \alpha} = $

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