If cos A-mcos B and cot A-B-2-tanB-A-2 and - is

The correct option is D cosA/cosB=m 1⇒cosA+cosB/cosA cosB=m+1/m 1 =2cos(A+B/2)cos(B A/2)/2sin(A+B/2)sin(B A/2)=m+1/m 1 ⇒ cot(A+B/2)=(m+1/m 1)tan(B A/2) ...

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Standard XII

Mathematics

Derivative Operator dy/dx

If cos A=mcos...

Question

If cos A= mcos B and cot $\frac{(A + B)}{2} = λ t a n (\frac{B - A}{2})$ and $λ$ is

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

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

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

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

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Solution

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

$\frac{c o s A}{c o s B} = \frac{m}{1} \Rightarrow \frac{c o s A + c o s B}{c o s A - c o s B} = \frac{m + 1}{m - 1}$
$= \frac{2 c o s (\frac{A + B}{2}) c o s (\frac{B - A}{2})}{2 s i n (\frac{A + B}{2}) s i n (\frac{B - A}{2})} = \frac{m + 1}{m - 1}$
$\Rightarrow c o t (\frac{A + B}{2}) = (\frac{m + 1}{m - 1}) t a n (\frac{B - A}{2})$

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If cos A= mcos B and cot(A+B)2=λtan(B−A2) and λ is

The correct option is C cosAcosB=m1⇒cosA+cosBcosA−cosB=m+1m−1 =2cos(A+B2)cos(B−A2)2sin(A+B2)sin(B−A2)=m+1m−1 ⇒cot(A+B2)=(m+1m−1)tan(B−A2)

If cos A= mcos B and cot $\frac{(A + B)}{2} = λ t a n (\frac{B - A}{2})$ and $λ$ is