If cos A = m cos B, then | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

If $90^\circ < A < 180^\circ$ and $\sin A = \frac{4}{5}$,then $\tan \frac{A}{2}$ is equal to

$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $

The value of $\log \tan 1^{\circ} + \log \tan 2^{\circ} + \log \tan 3^{\circ} + \ldots + \log \tan 89^{\circ}$ is equal to

$\tan 1^\circ \tan 2^\circ \tan 3^\circ \tan 4^\circ \dots \tan 89^\circ = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module