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If tan^2Atan^2B+tan^Btan^2C+tan^2Ctan^2A+2tan^2Atan^2Btan^C=1 then the value of sin^2A+sin^2B+sin^C=?

If tan^2Atan^2B+tan^Btan^2C+tan^2Ctan^2A+2tan^2Atan^2Btan^C=1 then the value of sin^2A+sin^2B+sin^C=?

Grade:12th pass

1 Answers

Arun
25750 Points
5 years ago
First we convert the given equation into sin and cosine functions.

⇒ [(sina)^2.(sinb)^2]/[(cosa)^2. (cosb)^2] + [(sinb)^2.(sinc)^2]/[(cosb)^2. (cosc)^2]
  + [(sina)^2.(sinc)^2]/[(cosa)^2. (cosc)^2] + 2[(sina)^2.(sinb)^2.(sinc)^2]/[(cosa)^2.      (cosb)^2.(cosc)^2]= 1

We know that (cosx)^2 = 1- (sinx)^2
Let sina = A
Let sinb = B
Let sinc= C

Hence,
(cosa)^2 = 1-A^2
(cosb)^2 = 1-B^2
(cosc)^2 = 1-C^2

Substituting this into the above equation, we get

(A^2)(B^2)/(1-A^2)(1-B^2) + (B^2)(C^2)/(1-B^2)(1-C^2) + (A^2)(C^2)/(1-A^2)(1-C^2) +2(A^2)(B^2)(C^2)/(1-A^2)(1-B^2)(1-C^2) = 1 

Multiplying the equation by (1-A^2)(1-B^2)(1-C^2), we get

(A^2)(B^2)(1-C^2) + (B^2)(C^2)(1-A^2) + (A^2)(C^2)(1-B^2) + 2(A^2)(B^2)(C^2) = 
  (1-A^2)(1-B^2)(1-C^2)

For simplicity, I will solve the left hand side and the right hand side separately and then equate.

LHS
(A^2)(B^2)(1-C^2) + (B^2)(C^2)(1-A^2) + (A^2)(C^2)(1-B^2) + 2(A^2)(B^2)(C^2)

=(A^2)(B^2) - (A^2)(B^2)(C^2) + (B^2)(C^2) - (A^2)(B^2)(C^2) + (A^2)(C^2) - (A^2)(B^2)(C^2) +2(A^2)(B^2)(C^2)

= (A^2)(B^2) + (B^2)(C^2) + (A^2)(C^2) - (A^2)(B^2)(C^2)

RHS
(1-A^2)(1-B^2)(1-C^2)

=(1-A^2)[1 - B^2 -C^2 +(B^2)(C^2)]

=1 - B^2 -C^2 +(B^2)(C^2) + (A^2)(B^2) +(A^2)(C^2) + (A^2)(B^2)(C^2)

LHS=RHS

⇒ (A^2)(B^2) + (B^2)(C^2) + (A^2)(C^2) - (A^2)(B^2)(C^2) 
         = 1 - B^2 -C^2 +(B^2)(C^2) + A^2 + (A^2)(B^2) +(A^2)(C^2) - (A^2)(B^2)(C^2)

Many of the terms get cancelled, which leaves us with

0 = 1 -A^2 -B^2 -C^2
⇒ A^2 + B^2 +C^2 = 1
⇒ (sina)^2 +(sinb)^2 +(sinc)^2 = 1

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