Question

09. If Sec - tan is a root of 3x2 + 5x + 1 = 0, then which of the following equation has the root seco + tan
O (A) x2 + 3x + 5 = 0​

Answer 1

Step-by-step explanation:

Given that:

*question is incomplete

Complete Question: If (secA- tanA) and(cosecA-cotA) are roots of 3x² + 5x + 1 = 0, then find the equation having roots (secA+ tanA) and(cosecA+cotA).

To find: Equation having roots (secA+ tanA) and(cosecA+cotA).

Solution:

Let

$secA - tanA = \alpha \\ \\ then \\ \\ secA + tanA = \frac{1}{ \alpha }$

other root

$cosecA - cotA= \beta \\ \\ then \\ \\ cosecA + cotA= \frac{1}{ \beta }$

Equation 3x²+5x+1 has roots (secA- tanA) and(cosecA-cotA)

then the equation having roots (secA+tanA) and(cosecA+cotA) is

$3 {\bigg( \frac{1}{x}\bigg )}^{2} + 5\bigg( \frac{1}{x}\bigg ) + 1 = 0 \\ \\ \frac{3}{ {x}^{2} } + \frac{5}{x} + 1 = 0 \\ \\ solve \\ \\ \frac{3 + 5x + {x}^{2} }{ {x}^{2} } = 0 \\ \\ {x}^{2} + 5x + 3 = 0$

Thus,

Equation is

$\bold{\red{ {x}^{2} + 5x + 3=0}}$

having roots (secA+tanA) and(cosecA+cotA)

Hope it helps you.