Question

if cos(a-b)=3÷5 and tan Ato b=2 then​

Answer 1

Answer:

$\bf \huge\pink{Given:}$

$\tan(a). \tan(b) = 2$

$or \: \frac{ \sin(A). \sin(B) }{ \cos(A) \cos(B) } = 2$

$or \: \sin(A) \sin(B) = \cos(A) \cos(B) .......(1)$

$and \: \cos(A + B) = \frac{3}{5}$

$and \: \cos(A + B) = \cos(A) \cos(B) + \sin(A) \sin(B)$

$or \: \cos(A) \cos(B) + \sin(A) \sin(B) = \frac{3}{5} .....(2)$

$From \: (1) \: we \: get$

$3 \cos(A) \cos(B) = \frac{3}{5}$

$Or \: \cos(A) \cos(B) = \frac{1}{5} ........(3)$

$So \: \sin(A). \sin(B) = 2 \times \frac{1}{5} = \frac{2}{5} .........(4)$

$now \: \cos(A + B) = \cos(A) \cos(B) - \sin(A)\sin(B)$

$Now \: from \: eqn.(3) \: and \: (4) \: we \: can \: say$

$\cos(A + B) = \frac{1}{5} - \frac{2}{5}$

$= \frac{1 - 2}{5}$

$= \frac{ - 1}{5}$