CBSE Class 10 Answered
trignometry
Asked by chiku999 | 30 Nov, 2009, 02:22: PM
Expert Answer
sinA/sinB=p
sin2A/sin2B = p2
sin2A/(1 - cos2B) = p2
But cos2B = cos2A/q2
sin2A/(1 - cos2A/q2) = p2
q2sin2A/(q2 - cos2A) = p2
q2sin2A = p2(q2 - cos2A) = p2q2 - p2 + p2sin2A
sin2A = (p2q2 - p2)/(q2-p2)
and hence cos2A = (q2 - p2q2)/(q2-p2)
tan2A = (p2q2 - p2)/(q2 - p2q2)
tan A = (±p/q)(q2-1)1/2/(1-p2)1/2
and we can see that tanA/tanB = p/q, hence
tan B = ±(q2-1)1/2/(1-p2)1/2
Regards,
Team,
TopperLearning.
Answered by | 30 Nov, 2009, 03:26: PM
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