If $α, β$ are solutions of ${sin}^{2} x + a sin x + b = 0$ and ${cos}^{2} x + c cos x + d = 0$ then $sin (α + β)$
equals

\frac{2 a c}{a^{2} + c^{2}}

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\frac{a^{2} + c^{2}}{2 a c}

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\frac{2 b d}{b^{2} + d^{2}}

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\frac{b^{2} + d^{2}}{2 b d}

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Solution

The correct option is A $\frac{2 a c}{a^{2} + c^{2}}$
$sin α + sin β = - a$
$cos α + cos β = - c$
$a^{2} + c^{2} = 1 + 1 + 2 (sin α sin β + cos α cos β) = 2 + 2 cos (α - β)$
$a c = sin α cos α + sin α cos β + sin β cos α + sin β cos β = 2 sin (α + β) [1 + cos (α - β)]$
$∴ \frac{2 a c}{a^{2} + c^{2}} = sin (α + β)$

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α, β

{sin}^{2} x + a sin x + b = 0

{cos}^{2} x + c cos x + d = 0

sin (α + β)

∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & b & c & d & e & f f & a & b & c & d & e e & f & a & b & c & d d & e & f & a & b & c c & d & e & f & a & b b & c & d & e & f & a \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = ∣ ∣ ∣ ∣ \begin{matrix} A & B & C C & A & B B & C & A \end{matrix} ∣ ∣ ∣ ∣

A = a^{2} - d^{2} + 2 c e - 2 b f

B = e^{2} - b^{2} + 2 a c - 2 d f

C = c^{2} - f^{2} + 2 a e - 2 b d

If $α$ and $β$ are solutions of $s i n^{2} x$ + a sin x + b = 0 as well as that of $c o s^{2} x$ + c cos x + d = 0, then sin( $α$ + $β$ ) is equal to

a x^{2} + b x + c = 0,

If a, b, c, d are in G.p., prove that :

(i) $(a^{2} + b^{2}), (b^{2} + c^{2}), (c^{2} + d)^{2}$ are in G.P.

(ii) $(a^{2} - b^{2}), (b^{2} - c^{2}), (c^{2} - d)^{2}$ are in G.P.

(iii) $\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}$ are in G.P.

(iv) $(a^{2} + b^{2} + c^{2}), (a b + b c + c d), (b^{2} + c^{2} + d^{2})$

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