In a triangle ABC the value of $∠ A$ is given by $5 c o s A + 3 = 0$ , then the equation whose roots are sin A and tan A will be

$15 x^{2} - 8 \sqrt{2} x + 16 = 0$

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$15 x^{2} - 8 x - 16 = 0$

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$15 x^{2} + 8 x - 16 = 0$

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$15 x^{2} - 8 x + 16 = 0$

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Solution

The correct option is C
$15 x^{2} + 8 x - 16 = 0$

Given that $5 c o s A + 3 = 0$ or $c o s A = - \frac{3}{5}$

Let $α = s i n A a n d β = t a n A,$ then the sum of roots

$= α + β = s i n A + t a n A = s i n A + \frac{s i n A}{c o s A} = \frac{s i n A}{c o s A} (1 + c o s A)$

$= \frac{\sqrt{1 - \frac{9}{25}}}{- \frac{3}{5}} (1 - \frac{3}{5}) = \frac{4}{- 5} . \frac{5}{3} . \frac{2}{5} = \frac{8}{- 15}$

and product of roots $α . β$ = sin A tan A $= \frac{s i n^{2} A}{c o s A}$

$= \frac{\frac{16}{25}}{- \frac{3}{15}} = - \frac{16}{25} \times \frac{5}{3} = - \frac{16}{15}$

Thus required equation is $x^{2} - (α + β) x + α β = 0$

$\Rightarrow x^{2} + \frac{8 x}{15} - \frac{16}{15} = 0 \Rightarrow 15 x^{2} + 8 x - 16 = 0$

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