Question

Find a polynomial whose zeroes are Tan 45º and Sec 60°.​

Answer 1

The quadratic polynomial is $x^{2}-3x+2.$

Step-by-step explanation:

Let the zeroes of polynomial = α and β

Given,

$\alpha=\tan 45$ and

$\beta=\sec 60$

To find, the polynomial = ?

∴ $\alpha=\tan 45$

⇒ $\alpha=1$ [ ∵ $\tan 45=1$]

Also, $\beta=\sec 60$ [ ∵ $\sec 60=2$]

⇒ $\beta=2$

∴ $\alpha=1$ and $\beta=2$

The quadratic polynomial is:

$x^{2} -(\alpha+\beta)x+\alpha.\beta$

$=x^{2} -(1+2)x+1.2$

$=x^{2}-3x+2$

Hence, the quadratic polynomial is $x^{2}-3x+2.$

Answer 2

A polynomial whose zeroes are Tan 45° and sec 60° is $x^{2}-3x+2$

• Given, Tan 45° and sec 60° are roots of the polynomial.
• If a, b are zeroes of a polynomial, then the polynomial can be written as

                 (x - a)(x - b)

• Here,

                 a =  Tan 45°,     b = sec 60°

                 a = 1                   b = 2

• Then the polynomial is

                ⇒ (x - 1)(x - 2)

                ⇒ $x^{2}-2x-x+2$

                ⇒ $x^{2}-3x+2$