Solveeit Logo
Login

Question

Question: If x<sub>1</sub>, x<sub>2</sub>, x<sub>3</sub>, x<sub>4</sub> are roots of the equation x<sup>4</sup...

If x1, x2, x3, x4 are roots of the equation x4 – x3 sin 2b + x2cos 2b – x cosb – sin b = 0 then ∑i=14tan⁡−1\sum_{i = 1}^{4}\tan^{- 1}xi is equal to

A

p – b

B

p – 2b

C

p/2 – b

D

p/2 – 2b

Answer

p/2 – b

Explanation

Solution

We have

S1 = S x1 = sin 2b

S2 = S x1x2 = cos 2b

S3 = S x1x2x3 = cos b

S4 = x1x2x3x4 = – sin b

So that ∑i=14tan⁡−1\sum_{i = 1}^{4}\tan^{- 1} xi = tan–1 S1−S31−S2+S4\frac{S_{1} - S_{3}}{1 - S_{2} + S_{4}}

= tan–1 sin⁡2β−cos⁡β1−cos⁡2β−sin⁡β\frac{\sin 2\beta - \cos\beta}{1 - \cos 2\beta - \sin\beta}

= tan–1 cos⁡β(2sin⁡⥂⥂β−1)sin⁡β(2sin⁡β−1)\frac{\cos\beta(2\sin ⥂ ⥂ \beta - 1)}{\sin\beta(2\sin\beta - 1)}

= tan cot b = tan–1 (tan (p/2 – b))

= p/2 – b