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If A,B,C are angles of a triangle such that Sin2A+Sin2B+Sin2c= constant. Then show that dA/dB= tan B – tan C/tan C -tan A.

ravi , 7 Years ago
Grade 12th pass
anser 2 Answers
Aditya Gupta
A+B+C= const
or dA+dB+dC= 0
and Sin2A+Sin2B+Sin2c= constant
differentiate and substitute dc=  – (dA+dB)
we get
dA/dB= (sin2B- sin2C)/(sin2C- sin2A)
but (sin2B- sin2C)/(sin2C- sin2A)= (tan B – tan C)/(tan C -tan A) [ you can prove this by using sinx-siny formula and writing tanx=sinx/cosx]
 
Last Activity: 7 Years ago
Manjot kaur
If A, B, C are angles of a triangle such that sin^2A+sin^2B+sin^2C =constant prove that dA/dB=tanC-tanB/tanA-tanC
 
Last Activity: 5 Years ago
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