If cos A = m cos B, then | vedclass

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Question

(A) Given that $\cos A = m \cos B,$ which implies $\frac{m}{1} = \frac{\cos A}{\cos B}.$Applying the componendo and dividendo rule:$\frac{m + 1}{m - 1

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given that $\cos A = m \cos B,$ which implies $\frac{m}{1} = \frac{\cos A}{\cos B}.$Applying the componendo and dividendo rule:$\frac{m + 1}{m - 1} = \frac{\cos A + \cos B}{\cos A - \cos B}.$Using the sum-to-product formulas:$\cos A + \cos B = 2 \cos \left( \frac{A + B}{2} ight) \cos \left( \frac{A - B}{2} ight)$$\cos A - \cos B = -2 \sin \left( \frac{A + B}{2} ight) \sin \left( \frac{A - B}{2} ight) = 2 \sin \left( \frac{A + B}{2} ight) \sin \left( \frac{B - A}{2} ight).$Substituting these into the expression:$\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{B - A}{2} ight)} = \cot \left( \frac{A + B}{2} ight) \cdot \frac{\cos \left( \frac{A - B}{2} ight)}{\sin \left( \frac{B - A}{2} ight)}.$Since $\cos \left( \frac{A - B}{2} ight) = \cos \left( \frac{B - A}{2} ight),$ we get:$\frac{m + 1}{m - 1} = \cot \left( \frac{A + B}{2} ight) \cot \left( \frac{B - A}{2} ight).$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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