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STD 11 - 3. trignometrical ratios,functions and identities — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 3. trignometrical ratios,functions and identities4 Marks
MCQ
If $\cos A = m\cos B,$ then
  • ✓
    $\cot \frac{{A + B}}{2} = \frac{{m + 1}}{{m - 1}}\tan \frac{{B - A}}{2}$
  • B
    $\tan \frac{{A + B}}{2} = \frac{{m + 1}}{{m - 1}}\cot \frac{{B - A}}{2}$
  • C
    $\cot \frac{{A + B}}{2} = \frac{{m + 1}}{{m - 1}}\tan \frac{{A - B}}{2}$
  • D
    None of these
✓

Answer

Correct option: A.
$\cot \frac{{A + B}}{2} = \frac{{m + 1}}{{m - 1}}\tan \frac{{B - A}}{2}$
a
(a) Given that $\cos A = m\,\,\cos B\, \Rightarrow \,\,\frac{m}{1} = \frac{{\cos A}}{{\cos B}}$

$ \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 \,\left( {\frac{{A + B}}{2}} \right)\,\cot \,\left( {\frac{{B - A}}{2}} \right)$

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

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