If cos A = m cos B, then | vedclass

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(A) Given $\cos A = m \cos B,$ we have $\frac{m}{1} = \frac{\cos A}{\cos B}.$Applying componendo and dividendo,we get $\frac{m + 1}{m - 1} = \frac{\co

Vedclass · Accepted Answer

$\cot \frac{A + B}{2} = \frac{m + 1}{m - 1} 	an \frac{B - A}{2}$

Vedclass · Answer

(A) Given $\cos A = m \cos B,$ we have $\frac{m}{1} = \frac{\cos A}{\cos B}.$Applying componendo and dividendo,we get $\frac{m + 1}{m - 1} = \frac{\cos A + \cos B}{\cos A - \cos B}.$Using sum-to-product formulas,$\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} ight) \cos \left( \frac{A - B}{2} ight)$ and $\cos A - \cos B = -2 \sin \left( \frac{A + B}{2} ight) \sin \left( \frac{A - B}{2} ight).$Thus,$\frac{m + 1}{m - 1} = \frac{2 \cos \left( \frac{A + B}{2} ight) \cos \left( \frac{A - B}{2} ight)}{-2 \sin \left( \frac{A + B}{2} ight) \sin \left( \frac{A - B}{2} ight)} = -\cot \left( \frac{A + B}{2} ight) \cot \left( \frac{A - B}{2} ight).$Since $\cot \left( \frac{A - B}{2} ight) = \cot \left( -\frac{B - A}{2} ight) = -\cot \left( \frac{B - A}{2} ight) = -\frac{1}{	an \left( \frac{B - A}{2} ight)},$we have $\frac{m + 1}{m - 1} = \cot \left( \frac{A + B}{2} ight) \frac{1}{	an \left( \frac{B - A}{2} ight)}.$Therefore,$\cot \left( \frac{A + B}{2} ight) = \frac{m + 1}{m - 1} 	an \left( \frac{B - A}{2} ight).$

If $\cos A = m \cos B,$ then

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