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Transformation Formulae — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTransformation Formulae1 Mark
Question
If $\cos\text{A}=\text{m}\cos\text{B},$ than write the value of $\cot\frac{\text{A+B}}{2}\cot\frac{\text{A}-\text{B}}{2}.$
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Answer

We have,
$\cos\text{A}=\text{m}\cos\text{B}$
$\frac{\cos\text{A}}{\cos\text{B}}=\text{m}$
$\Rightarrow\ \frac{\cos\text{A}}{\cos\text{B}}+1=\text{m}+1$
$\Rightarrow\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}}=\text{m}+1...(\text{i})$
Again,
$\frac{\cos\text{A}}{\cos\text{B}}=\text{m}$
$\Rightarrow\ \frac{\cos\text{A}}{\cos\text{B}}-1=\text{m}-1$
$\Rightarrow\ \frac{\cos\text{A}-\cos\text{B}}{\cos\text{B}}=\text{m}-1...(\text{ii})$
Dividing equation (i) by equation (ii), we get
$\frac{\frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}}}{\frac{\cos\text{A}-\cos\text{B}}{\cos\text{B}}}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{A}-\cos\text{B}}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \frac{2\cos\Big(\frac{\text{A+B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{-2\sin\Big(\frac{\text{A+B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)}=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ -\cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{\text{m}+1}{\text{m}-1}$
$\Rightarrow\ \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{1+\text{m}}{1-\text{m}}$
$\therefore\ \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)=\frac{1+\text{m}}{1-\text{m}}$

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