1.

If acosθ + bsinθ = c have roots α and β. Then, what will be the value of sinα * sinβ ?(a) 2bc/(a^2 + b^2)(b) 0(c) 1(d) (c^2 + a^2)/(a^2 + b^2)The question was posed to me in exam.Enquiry is from Quadratic Equations in division Complex Numbers and Quadratic Equations of Mathematics – Class 11

Answer»

The correct OPTION is (d) (c^2 + a^2)/(a^2 + b^2)

For explanation I would say: Given, acosθ + bsinθ = c

So, this implies acosθ = c – bsinθ

Now squaring both the sides we get,

(acosθ)^2 = (c – bsinθ)^2

a^2 cos^2 θ = c^2 + b^2 sin^2 θ – 2b c sinθ

a^2 (1- sin^2 θ) = c^2 + b^2 sin^2 θ – 2b c sinθ

a^2 – a^2 sin^2 θ = c^2 + b^2 sin^2 θ – 2b c sinθ

Now rearranging the elements,

(a^2 + b^2) sin^2 θ – 2b c sinθ+( c^2 – a^2) = θ

So, as sum of the roots are in the formc/a if there is a QUADRATIC equation ax^2 + BX + c = 0

Now , we can CONCLUDE that

sinα + sinβ = (c^2 + a^2)/(a^2 + b^2).


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