If A=m B, then - Vidyadip

MCQ

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

✓
$\cot \left(\frac{ A + B }{2}\right)=\frac{ m +1}{m-1} \tan \left(\frac{B- A }{2}\right)$
B
$\tan \left(\frac{ A + B }{2}\right)=\frac{ m +1}{m-1} \cot \left(\frac{B- A }{2}\right)$
C
$\cot \left(\frac{A+B}{2}\right)=\frac{m+1}{m-1} \tan \left(\frac{A-B}{2}\right)$
D
None of these

✓

Answer

Correct option: A.

$\cot \left(\frac{ A + B }{2}\right)=\frac{ m +1}{m-1} \tan \left(\frac{B- A }{2}\right)$

(A)
Given that, $\cos A=m \cos B$
$\Rightarrow \frac{m}{1}=\frac{\cos A}{\cos B}$
By componendo and dividendo, we get
$\frac{m+1}{m-1}=\frac{\cos A+\cos B}{\cos A-\cos B}$
$=\frac{2 \cos \left(\frac{A+ B }{2}\right) \cos \left(\frac{ B - A }{2}\right)}{2 \sin \left(\frac{A+ B }{2}\right) \sin \left(\frac{ B - A }{2}\right)}$
$\ldots[\because \cos (B-A)=\cos (A-B)]$
$=\cot \left(\frac{ A + B }{2}\right) \cot \left(\frac{ B - A }{2}\right)$
$\Rightarrow \cot \left(\frac{ A + B }{2}\right)=\frac{ m +1}{m-1} \tan \left(\frac{B- A }{2}\right)$

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