Question

Let α, β are roots of the equation sin²x + a sin x + b = 0 and also of cos²x + c cos x + d = 0.

If cos (α + β) = $\frac{a_{0}a^{2} + a_{1}b^{2} + a_{2}c^{2} + a_{3}d^{2}}{b_{0}a^{2} + b_{1}b^{2} + b_{2}c^{2} + b_{3}d^{2}}$, then

$\frac{- a_{0} + a_{1} + a_{2} + a_{3}}{b_{0} + b_{1} + b_{2} + b_{3}}$ equals

• A. 0
• B. – 1
• C. 1
• D. $\frac{- a_{0}a_{1}a_{2}a_{3}}{b_{0}b_{1}b_{2}b_{3}}$

Answer 1

Answer: (C)

1

Answer 2

sin α + sin β = – a; sin α sin β = b

cos α + cos β = – c; cos α cos β = d

2 sin $\left( \frac{\alpha + \beta}{2} \right)$cos$\left( \frac{\alpha - \beta}{2} \right)$ = – a

2 cos $\left( \frac{\alpha + \beta}{2} \right)$ cos$\left( \frac{\alpha - \beta}{2} \right)$ = – c

tan $\left( \frac{\alpha + \beta}{2} \right)$= $\frac{a}{c}$

$\frac{1 - \frac{a^{2}}{c^{2}}}{1 + \frac{a^{2}}{c^{2}}}$= $\frac{a_{0}a^{2} + a_{1}b^{2} + a_{2}c^{2} + a_{3}d^{2}}{b_{0}a^{2} + b_{1}b^{2} + b_{2}c^{2} + b_{3}d^{2}}$

⇒ a₀ = – 1, a₁ = 0, a₂ = 1, a₃ = 0

b₀ = 1, b₁ = 0, b₂ = 1, b₃ = 0

⇒ $\frac{- a_{0} + a_{1} + a_{2} + a_{3}}{b_{0} + b_{1} + b_{2} + b_{3}}$= 1