cos Ap,sin AIfsin Bcos B-q, find tan A and tan B.

1.	cos Ap,sin AIfsin Bcos B-q, find tan A and tan B.
Answer» Given , sinA/sinB=p and cosA/cosB=q So sinA= psinB and cosA=qcosB............(1) Now dividing above two equation we get, tan A= p/q* tanBor tan B= q/ptanA ........(2) Now multiplying both equation in (1) we get, sinAcosA= pqsinBcosB Now dividing by cos2^Acos^2B we get, sinAcosA/cos^2Acos^2B=pqsinBcosB/cos^2Acos^2B sec^2BtanA = pq se^c2AtanB(1+tan2B)tanA=pq (1+tan2A) tanB Now from equation (2) we get, (1+(q/ptanA)^2)tanA=pq (1+tan^2A) q/ptanA 1+q^2/p^2tan^2A=q^2+q^2tan^2A1−q^2=(q^2−q^2/p^2)tan^2A So,tan^2A= p^2(1−q^2)/q^2(p^2−1) tan A=p/q sqrt[(1−q^2)/(p^2−1)]

Answer»

Given , sinA/sinB=p and cosA/cosB=q

So sinA= psinB and cosA=qcosB............(1)

Now dividing above two equation we get,

tan A= p/q* tanBor tan B= q/p*tanA ........(2)

Now multiplying both equation in (1) we get,

sinAcosA= pqsinBcosB

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

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

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

Now from equation (2) we get,

(1+(q/ptanA)^2)tanA=pq (1+tan^2A) q/p*tanA

1+q^2/p^2*tan^2A=q^2+q^2tan^2A1−q^2=(q^2−q^2/p^2)tan^2A

So,tan^2A= p^2(1−q^2)/q^2(p^2−1)

tan A=p/q sqrt[(1−q^2)/(p^2−1)]

Discussion