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General Solution of tan theta = tan alpha

If α and ...

Question

If $α$ and $β$ are the solutions of $a cos θ + b sin θ = c$ , then show that
$cos α + cos β = 2 a c / (a^{2} + b^{2})$
$cos α \times cos β = (c^{2} - b^{2}) / (a^{2} + b^{2})$

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Solution

$a cos θ + b sin θ = c$
$c - a cos θ = b sin θ$
$(c - a cos θ)^{2} = b^{2} {sin}^{2} θ$
$c^{2} + a^{2} {cos}^{2} θ - 2 a c cos θ = b^{2} - b^{2} {cos}^{2} θ$
$(a^{2} + b^{2}) {cos}^{2} θ - 2 a c cos θ + (c^{2} - b^{2}) = 0$
$⟹ cos α + cos β = \frac{2 a c}{(a^{2} + b^{2})}$
$⟹ cos α \times cos β = (c^{2} - b^{2}) / (a^{2} + b^{2})$

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General Solution of tan theta = tan alpha

Standard XI Mathematics

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