Let and be the roots of the quadratic equation x^2 sin ⁡ -x ( sin ⁡ cos ⁡ +1 )+ cos ⁡ =0(0

Question

Let and be the roots of the quadratic equation x^2 sin ⁡ -x ( sin ⁡ cos ⁡ +1 )+ cos ⁡ =0(0< <45^ ) and < . Then 0.05em( ^n+ (-1 )^n ^n) is equal to :

Accepted Answer

Correct answer: C. (c) JEE Main 2019 x^2 sin ⁡ -x ( sin ⁡ cos ⁡ +1 )+ cos ⁡ =0 x^2 sin ⁡ -x sin ⁡ cos ⁡ -x+ cos ⁡ =0 x sin ⁡ - (x- cos ⁡ )-1 (x- cos ⁡ )=0 (x- cos ⁡ ) (x sin ⁡ -1 )=0 ∴x= cos ⁡ , cosec ⁡ , ∈ (0 ,45^ ) ∴ = cos ⁡ , = cosec ⁡ 0.05em ^n=1+ cos ⁡ + cos^2 ⁡ + … = 11- cos ⁡ 0.05em (-1 )^n ^n=1- 1 cosec + 1 cosec^2 - 1 cosec^3 + … =1- sin ⁡ + sin^2 ⁡ - sin^3 ⁡ + … = 11+ sin ⁡ ∴ 0.05em( ^n+ (-1 )^n ^n)= 0.05em ^n+ 0.05em (-1 )^n ^n = 11- cos ⁡ + 11+ sin ⁡

$Let α and β be the roots of the quadratic equation x^{2} sin θ - x (sin θ cos θ + 1) + cos θ = 0 (0 < θ < 45^{\circ}) and α < β . Then \sum (α^{n} + \frac{( - 1 ) ^{n}}{β ^{n}}) is equal to :$

Solution

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