If cot α cot β =2 , then ( cos(α +β)/ cos(α-β)) is equal to

Question

If cot α cot β =2 , then ( cos(α +β)/ cos(α-β)) is equal to (A) 3 (B)  (2/3)  (C)  (1/3)  (D)  tan α tan β

Tardigrade Admin · Accepted Answer

( cos(α +β)/ cos(α-β)) = ( cosα cos β - sin α sin β / cos α cos β + sin α sin β) Dividing numerator and denominator by sin α sin β ( cot α cot β-1/ cot α cot β + 1) = (2-1/2+1)=(1/3)

Q. If $cot α cot β = 2$ , then $\frac{c o s ( α + β )}{c o s ( α - β )}$ is equal to

3

$\frac{2}{3}$

$\frac{1}{3}$

$tan α tan β$

$\frac{c o s ( α + β )}{c o s ( α - β )} = \frac{c o s α c o s β - s i n α s i n β}{c o s α c o s β + s i n α s i n β}$
Dividing numerator and denominator by $sin α sin β$
$\frac{c o t α c o t β - 1}{c o t α c o t β + 1} = \frac{2 - 1}{2 + 1} = \frac{1}{3}$

Q. If cotαcotβ=2 , then cos(α−β)cos(α+β)​ is equal to

3

32​

31​

tanαtanβ

cos(α−β)cos(α+β)​=cosαcosβ+sinαsinβcosαcosβ−sinαsinβ​ Dividing numerator and denominator by sinαsinβ cotαcotβ+1cotαcotβ−1​=2+12−1​=31​

Q. If $cot α cot β = 2$ , then $\frac{c o s ( α + β )}{c o s ( α - β )}$ is equal to

$\frac{2}{3}$

$\frac{1}{3}$

$tan α tan β$

$\frac{c o s ( α + β )}{c o s ( α - β )} = \frac{c o s α c o s β - s i n α s i n β}{c o s α c o s β + s i n α s i n β}$
Dividing numerator and denominator by $sin α sin β$
$\frac{c o t α c o t β - 1}{c o t α c o t β + 1} = \frac{2 - 1}{2 + 1} = \frac{1}{3}$