Question

find polynomial for whose zeros are tan square 45 degrees and sec 60°​

Answer 1

Answer:

The required polynomial is x² - 3x + 2.

Given:

• The zeroes of the polynomial are tan²45° and sec 60°

To find:

• The polynomial.

Solution:

=> tan²45° = 1² = 1

=> sec 60° = 2

Sum of zeroes = tan²45° + sec 60°

Sum of zeroes = 1 + 2

Sum of zeroes = 3...(1)

Product of zeroes = tan²45° sec 60°

Product of zeroes = 1 × 2

Product of zeroes = 2...(2)

Quadratic polynomial can be written as

p(x) = x² - (Sum of zeroes)x + (Product of zeroes)

=> p(x) = x² - 3x + 2

Therefore, the required polynomial is

x² - 3x + 2.

Answer 2

$\large{\boxed{\bf{AnswEr}}}$

• The polynomial is x² - 3x + 2

Given :-

• Zeroes of a polynomial as tan²45° and sec60°.

To Find :-

• The polynomial

__________________________

$\large{\boxed{\bf{Solution}}}$

• tan²45° ➪ (1)² ➪ 1

• sec 60° ➪ 2

★ Zeroes of the polynomial are 1 and 2 respectively.

︎︎︎ Sum of zeroes

➪ 1 + 2

➪ 3

︎︎︎ Product of zeroes

➪ 1 × 2

➪ 2

Formula of quadratic polynomial is :-

➪ x² - (sum of zeroes)x + product of zeroes

➪ x² - (3)x + 2

➪ x² - 3x + 2

Hence, the quadratic polynomial is

x² - 3x + 2