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# Please solve this as soon as possible. Thank you. Regards.

Aarushi Ahlawat
41 Points
2 years ago
Group terms having ‘a’ and ‘b’ and then convert tan to sin and cos. Solve ans eliminate common terms.
$a~tan(\alpha) + b~tan (\beta) = (a+b) tan \left ( \frac{\alpha + \beta}{2} \right)$

$a~tan(\alpha) -~a tan \left ( \frac{\alpha + \beta}{2} \right) = b~tan \left ( \frac{\alpha + \beta}{2} \right) -b~tan (\beta)$
$a\left ( tan(\alpha) -tan \left ( \frac{\alpha + \beta}{2} \right) \right )= b\left ( tan \left ( \frac{\alpha + \beta}{2} \right) -tan (\beta) \right )$
$a\left ( \frac{sin(\alpha)}{cos(\alpha)} - \frac{sin \left ( \frac{\alpha + \beta}{2} \right) }{cos \left ( \frac{\alpha + \beta}{2} \right)} \right) = b\left ( \frac{sin \left ( \frac{\alpha + \beta}{2} \right) }{cos \left ( \frac{\alpha + \beta}{2} \right)} - \frac{sin(\beta)}{cos(\beta)} - \right)$
$a\left ( \frac{sin(\alpha) cos \left ( \frac{\alpha + \beta}{2}\right) - cos(\alpha) sin \left ( \frac{\alpha + \beta}{2}\right)}{cos(\alpha)cos \left ( \frac{\alpha + \beta}{2} \right)} \right) = b\left ( \frac{sin \left ( \frac{\alpha + \beta}{2}\right)cos(\beta) - cos \left ( \frac{\alpha + \beta}{2}\right)sin(\alpha)}{cos(\beta)cos \left ( \frac{\alpha + \beta}{2} \right)}\right)$
$a\left ( \frac{sin \left (\alpha- \frac{\alpha + \beta}{2}\right)}{cos(\alpha)} \right) = b\left ( \frac{sin \left ( \frac{\alpha + \beta}{2}-\beta\right)}{cos(\beta)}\right)$

$a\left ( \frac{sin \left (\frac{\alpha - \beta}{2}\right)}{cos(\alpha)} \right) = b\left ( \frac{sin \left ( \frac{\alpha - \beta}{2}\right)}{cos(\beta)}\right)$

$\frac{cos(\alpha)}{cos(\beta)} = \frac{a}{b}$