Question

sin A÷Sin B = m and Cos A ÷cos B = n and find the value of tanB

Answer 1

Given , sinA/sinB=m and cosA/cosB=n

So sinA= msinB and cosA=ncosB............(1)

Now dividing above two equation we get,

tan A= m/n* tanB
or tan B= n/m*tanA ........(2)

Now multiplying both equation in (1) we get,

sinAcosA= mnsinBcosB

Now dividing by cos2^Acos^2B we get,

sinAcosA/cos^2Acos^2B=mnsinBcosB/cos^2Acos^2B

sec^2B*tanA = mn se^c2AtanB
(1+tan2B)tanA=mn (1+tan2A) tanB

Now from equation (2) we get,

(1+(n/mtanA)^2)tanA=mn (1+tan^2A) n/m*tanA

1+n^2/m^2*tan^2A=n^2+n^2tan^2A
1−n^2=(n^2−n^2/m^2)tan^2A

So,
tan^2A= m^2(1−n^2)/n^2(m^2−1)

tan A=m/n sqrt[(1−n^2)/(m^2−1)]

this is your answer.....