If \( \cos A = m\cos B \), then \( \cot\frac{A+B}{2}\cot\frac{B-A}{2} \) is
Options:
(a) \( \frac{m-1}{m+1} \)
(b) \( \frac{m+2}{m-2} \)
(c) \( \frac{m+1}{m-1} \)
(d) none of these
Solution:
Given,
\[
\cos A=m\cos B
\]
\[
\frac{\cos A}{\cos B}=m
\]
Using identities,
\[
\cos A+\cos B
=
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
and
\[
\cos A-\cos B
=
-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
\]
\[
m
=
\frac{\cos A+\cos B+\cos A-\cos B}
{\cos A+\cos B-(\cos A-\cos B)}
\]
\[
=
\frac{
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
–
2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
}
{
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
+
2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
}
\]
Dividing numerator and denominator by
\[
2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
\]
\[
m
=
\frac{
\cot\frac{A+B}{2}\cot\frac{A-B}{2}-1
}
{
\cot\frac{A+B}{2}\cot\frac{A-B}{2}+1
}
\]
Let
\[
x=\cot\frac{A+B}{2}\cot\frac{A-B}{2}
\]
\[
m=\frac{x-1}{x+1}
\]
\[
mx+m=x-1
\]
\[
x(m-1)=-(m+1)
\]
\[
x=\frac{m+1}{1-m}
\]
Since,
\[
\cot\frac{B-A}{2}=-\cot\frac{A-B}{2}
\]
therefore,
\[
\cot\frac{A+B}{2}\cot\frac{B-A}{2}
=
\frac{m+1}{m-1}
\]
\[
\boxed{\frac{m+1}{m-1}}
\]
Correct option: (c)