Question

if cosA=tanB, cosB=tanC, cosC=tanA, then sinA=?
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Answer 1

Answer:

√5 – 1/2

Step-by-step explanation:

Given  

if cosA=tanB, cosB=tanC, cosC=tanA, then sinA=?

Consider  

Cos A = tan B

Cos A = sinA / cosB

Squaring both sides we get

Cos^2 A cos^2 B = sin^2 B

(1 – sin^2 A)(1 – sin^2 B) = sin^2 B

Now let sin A = x and sin B = y, also sin C = z

So (1 –  x^2)(1 – y^2) = y^2

Now 1 – x^2 = y^2 / 1 – y^2

      X^2 = 1 - y^2 / 1- y^2

     X^2 = 1 – 2y^2 / 1 – y^2----------1

Similarly for y and z we get

       Y^2 = 1 – 2x^2 / 1 – z^2----------2

      Z^2 = 1 – 2 x^2 / 1 – x^2------------3

From 2 and 3 we get  

     Y^2 = 1 - 2(1 – 2x^2/1 – x^2) / 1 – (1 – 2x^2 / 1 – x^2)

      Y^2 = 1 – x^2 – 2(1 – 2 x^2) / 1 – x^2 – (1 – 2x^2)

     Y^2 = 3 x^2 – 1 / x^2

    Y^2 = 3 – 1/x^2-----------------4

 Now from 1 and 4 we have

  X^2 = 1 – 2y^2 /1 – y^2

 X^2 = 1 – 2(3 – 1/x^2) / 1 – (3 – 1/x^2)

 X^2 = 2/x^2 – 5 / 1 / x^2 – 2

X^2 = 2 – 5 x^2 / 1 – 2x^2

2 x^4 – 6 x^2 + 2 = 0

X^4 – 3 x^2 + 1 = 0

Using x = - b ± √b^2 – 4ac / 2a we get

X^2 = 3 ± √9 – 4 / 2

X^2 = 3 ± √5 / 2

Now sin^2 A ≥ 1

X^2 = 3 - √5 / 2

X = √3 - √5 / 2

Multiply and divide by 2 we get

X = √6 - 2√5 / 4

X = √(√5)^2 + 1^2 - 2√5 x 1 / 4

X = √5 – 1 / 2

So numerical value of sin A = √5 – 1/2