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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae1 Mark
MCQ
If $\cos\text{A}=\text{m}\cos\text{B},$ than $\cot\frac{\text{A+B}}{2}\cot\frac{\text{B}-\text{A}}{2}=$
  • A
    $\frac{\text{m}-1}{\text{m}+1}$
  • B
    $\frac{\text{m}+2}{\text{m}-2}$
  • ✓
    $\frac{\text{m}+1}{\text{m} -1}$
  • D
    None of these
✓

Answer

Correct option: C.
$\frac{\text{m}+1}{\text{m} -1}$
Given :
$\cos\text{A}=\text{m}\cos\text{B}$
$\Rightarrow\ \frac{\cos\text{A}}{\cos\text{B}}=\frac{\text{m}}{1}$
$\Rightarrow\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{A}-\cos\text{B}}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \frac{2\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)}{-2\sin\Big(\frac{\text{B+A}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$
$\Big[\because\ \cos\text{A}+\cos\text{B}=2\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A+B}}{2}\Big)$ and $ \cos\text{A}-\cos\text{B}=2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)\Big]$
$\Rightarrow\ \frac{\cos\Big(\frac{\text{B}-\text{A}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)}{\sin\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{B}-\text{A}}{2}\Big)=\frac{\text{m}+1}{\text{m}-1}$

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