lf $α, β$ 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 $\frac{2 a c}{a^{2} + c^{2}}$
Let $α$ and $β$ are roots of the given equation ${sin}^{2} x + a sin x + b = 0$ . Then,
$\Rightarrow$ $sin α + sin β = \frac{- a}{1} = - a$ .......... (A)

$\Rightarrow$ $sin α sin β = \frac{b}{1} = b$

Let $α$ and $β$ are roots of the given equation ${cos}^{2} x + c cos x + d = 0$ . Then,

$\Rightarrow$ $cos α + cos β = \frac{- c}{1} = - c$ .......... (B)

$\Rightarrow$ $cos α cos β = \frac{d}{1} = d$

$a^{2} + c^{2} = (sin α + sin β)^{2} + (cos α + cos β)^{2}$

$= {sin}^{2} α + {sin}^{2} β + 2 sin α sin β + {cos}^{2} α + {cos}^{2} β + 2 cos α cos β$

$= 1 + 1 + 2 (sin α sin β + cos α cos β)$

$= 2 [1 + cos (α - β)]$

$= 2 \times 2 {cos}^{2} (\frac{α - β}{2})$

$= 4 {cos}^{2} (\frac{α - β}{2})$

$∴$ $a^{2} + c^{2}$ $= 4 {cos}^{2} (\frac{α - β}{2})$ ---- ( 1 )

Now, from A and B
$2 sin \frac{α + β}{2} . cos \frac{α - β}{2} = - a$ --- ( 2 )

$2 cos \frac{α + β}{2} . cos \frac{α - β}{2} = - c$ ---- ( 3 )

Multiplying ( 2 ) and ( 3 ),

$\Rightarrow$ $4 sin \frac{α + β}{2} . cos \frac{α + β}{2} . {cos}^{2} (\frac{α - β}{2}) = a c$

$\Rightarrow$ $2 sin (α + β) (\frac{a^{2} + c^{2}}{4}) = a c$ [ From ( 1 ) ]

$\Rightarrow$ $sin (α + β) = \frac{2 a c}{a^{2} + c^{2}}$

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

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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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