If #tan (A-B) /tanA +sin^2C /sin^2A=1 #, prove that #tanA tanB =tan^2 C# ?

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Answer 1 · 2017-09-19T02:41:28

#tan (A-B) /tanA +sin^2C /sin^2A=1 #

#=>sin^2C /sin^2A=1-tan (A-B) /tanA #

#=>sin^2C /sin^2A=1-(sin(A-B) cosA)/(cos(A-B)sinA ) #

#=>sin^2C /sin^2A=(sinAcos(A-B)-sin(A-B) cosA)/(cos(A-B)sinA ) #

#=>sin^2C=sin(A-A+B)/(cos(A-B)sinA ) xxsin^2A#

#=>sin^2C=(sinAsinB)/cos(A-B)#

#=>csc^2C=cos(A-B)/(sinAsinB)#

#=>csc^2C=(cosAcosB+sinAsinB)/(sinAsinB)#

#=>csc^2C=(cosAcosB)/(sinAsinB)+(sinAsinB)/(sinAsinB)#

#=>1+cot^2C=cotAcotB+1#

#=>cotAcotB=cot^2C#

#=>tanAtanB=tan^2C#

Proved