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Question: If \(\cos A = m\cos B,\) then...

If cos⁡A=mcos⁡B,\cos A = m\cos B, then

A

cot⁡A+B2=m+1m−1tan⁡B−A2\cot\frac{A + B}{2} = \frac{m + 1}{m - 1}\tan\frac{B - A}{2}

B

tan⁡A+B2=m+1m−1cot⁡B−A2\tan\frac{A + B}{2} = \frac{m + 1}{m - 1}\cot\frac{B - A}{2}

C

cot⁡A+B2=m+1m−1tan⁡A−B2\cot\frac{A + B}{2} = \frac{m + 1}{m - 1}\tan\frac{A - B}{2}

D

None of these

Answer

cot⁡A+B2=m+1m−1tan⁡B−A2\cot\frac{A + B}{2} = \frac{m + 1}{m - 1}\tan\frac{B - A}{2}

Explanation

Solution

Given that cos⁡A=mcos⁡B⇒m1=cos⁡Acos⁡B\cos A = m\cos B \Rightarrow \frac{m}{1} = \frac{\cos A}{\cos B}

⇒m+1m−1=cos⁡A+cos⁡Bcos⁡A−cos⁡B=2cos⁡(A+B2)cos⁡(B−A2)2sin⁡(A+B2)sin⁡(B−A2)\Rightarrow \frac{m + 1}{m - 1} = \frac{\cos A + \cos B}{\cos A - \cos B} = \frac{2\cos\left( \frac{A + B}{2} \right)\cos\left( \frac{B - A}{2} \right)}{2\sin\left( \frac{A + B}{2} \right)\sin\left( \frac{B - A}{2} \right)}

=cot⁡(A+B2)cot⁡(B−A2)= \cot\left( \frac{A + B}{2} \right)\cot\left( \frac{B - A}{2} \right)

Hence, cot⁡(A+B2)=m+1m−1tan⁡B−A2\cot\left( \frac{A + B}{2} \right) = \frac{m + 1}{m - 1}\tan\frac{B - A}{2}.