if cosA=tanB.cosB=tanC,CosC=tant then sinA
Answers
Step-by-step explanation:
Ans:-√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