If cosA -mcos B- then A-B-2B-A-2-

The correct option is D m+1/m 1 Given: cosA=mcosB ⇒cosA/cosB=m/1 ⇒cosA+cosB/cosB cosB=m+1/m 1 ⇒2cos ( A B/2 )cos ( A+B/2 )/ 2sincos ( B+A/2 )sin ( B A/2 )=m+1/ ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If cosA =mcos...

Question

If $c o s A = m c o s B$ , then $c o t \frac{A + B}{2} c o t \frac{B - A}{2} =$

$\frac{m - 1}{m + 1}$

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$\frac{m + 2}{m - 2}$

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$\frac{m + 1}{m - 1}$

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None of these

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Solution

The correct option is C
$\frac{m + 1}{m - 1}$

Given:
$c o s A = m c o s B \Rightarrow \frac{c o s A}{c o s B} = \frac{m}{1} \Rightarrow \frac{c o s A + c o s B}{c o s B - c o s B} = \frac{m + 1}{m - 1} \Rightarrow \frac{2 c o s (\frac{A - B}{2}) c o s (\frac{A + B}{2})}{- 2 s i n c o s (\frac{B + A}{2}) s i n (\frac{B - A}{2})} = \frac{m + 1}{m - 1} ⎡ ⎢ ⎣ \begin{matrix} ∵ c o s A + c o s B = 2 c o s (\frac{A - B}{2}) c o s (\frac{A + B}{2}) a n d c o s A - c o s B = 2 s i n (\frac{A + B}{2}) c o s (\frac{B - A}{2}) \end{matrix} ⎤ ⎥ ⎦ \Rightarrow \frac{c o s (\frac{B - A}{2}) c o s (\frac{A + B}{2})}{s i n (\frac{A - B}{2}) s i n (\frac{B - a}{2})} = \frac{m + 1}{m - 1} \Rightarrow c o t (\frac{A + B}{2}) c o t (\frac{B - A}{2}) = \frac{m + 1}{m - 1}$

Suggest Corrections

If cosA=mcosB, then cotA+B2cotB−A2=

If $c o s A = m c o s B$ , then $c o t \frac{A + B}{2} c o t \frac{B - A}{2} =$