Question

PROVETHE FOLLOWING :

Answer 1

$We\;know\;that\\\\Cos\theta = 1 - 2Sin^2(\frac{\theta}{2})\\\\1 - Cos\theta = 2Sin^2(\frac{\theta}{2})\\\\In\;the\;similar\;way\\\\1 - Cos(A + B) = 2Sin^2(\frac{A + B}{2})\\\\We\;also\;know\;that\\\\CosB - CosA = 2Sin(\frac{A + B}{2})Sin(\frac{A - B}{2})\\\\CosA - CosB = -2Sin(\frac{A + B}{2})Sin(\frac{A - B}{2})$

$Substituting\;the\;above\;values\;in\;the\;given\;problem\;we\;get\;\\\\\frac{2Sin^2(\frac{A + B}{2}) + 2Sin(\frac{A + B}{2})Sin(\frac{A - B}{2})}{2Sin^2(\frac{A + B}{2}) - 2Sin(\frac{A + B}{2})Sin(\frac{A - B}{2})}\\\\\frac{Sin(\frac{A + B}{2}) + Sin(\frac{A - B}{2})}{Sin(\frac{A + B}{2}) - Sin(\frac{A - B}{2})}$

$We\;know\;that\\\\Sin(\frac{A + B}{2}) + Sin(\frac{A - B}{2}) = 2Sin(\frac{A}{2})Cos(\frac{B}{2})\\\\Sin(\frac{A + B}{2}) - Sin(\frac{A - B}{2}) = 2Cos(\frac{A}{2})Sin(\frac{B}{2})\\\\Substituting\;we\;get\\\\Tan(\frac{A}{2})Cot(\frac{B}{2})$