Asked by Rabin

Prove: 1+(TanA/SinB)^2/1+(TanA/SinC)^2 = 1+(SinA/TanB)^2/1+(SinA/TanC)^2
Answered by Neeraj
Solve this
Answered by Rahul sah
Excellent
Answered by Asmir Waiba
L.H.S =1+(TanA/SinB)²/ 1+(TanA/SinC)²
= 1+Tan²A.Cosec²B/ 1+Tan²A/Cosec²C
= 1+Tan²A(1+Cot²B)/1+Tan²A(1+Cot²C)

=1+Tan²A+Tan²A.Cot²B/
1+Tan²A+Tan²A.Cot²C

=Sec²A+Tan²A.Cot²B/ Sec²A+Tan²A.Cot²C

=[(1/Cos²A)+(Sin²A.Cot²B)/Cos²A]/
[(1/Cos²A)+(Sin²A.Cot²C)/Cos²A]

=[(1+Sin²A.Cot²B)/Cos²A]/
[(1+Sin²A.Cot²C)/Cos²A]

=(1+Sin²A.Cot²B)/ (1+Sin²A.Cot²C)
=(1+Sin²A/Tan²B)/ (1+Sin²A/Tan²C)
=1+(SinA/TanB)²/ 1+(SinA/TanC)²
=R.H.S
Answer
Sin²A+Cos²A=1
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