(i) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2
We know that sin 60° =Y hence the option (A) is correct.

(ii)(1 – tan² 45°)/(1 + tan² 45°)
Putting the value of each trigonometric ratios, we get
(1-(1)²)(1+(1)²) = (1-1)/(1+1) = 0/2 = 0
Hence, the option (D) is correct.

(iii) sin 2A = 2 sin A
We know that sin 0 = 0, hence, the option (A) is correct.

(iv) (2 tan 30°)/(1 – tan²30°)
Putting the value of each trigonometric ratios, we get
(2(1/√3))/(1-(1/√3)²) = (1/√3)/(1 – 1/3) = (2/√3)/(2/3) = 3/√3 = √3
We know that tan 60° = √3, hence, the option (C) is correct.

Here is the video explanation of the above question😁🤗

Let, p(x) = ax³ + bx² + cx + d be a cubic polynomial whose zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
Given that,
α + β + γ = 2
αβ + βγ + γα = -7
αβγ = -14
We know that,
α + β + γ = -(Cofficient of x²)/(Cofficient of x³)
αβ + βγ + γα = (Cofficient of x)/(Cofficient of x³)
αβγ = -(Constant term)/Cofficient of x³)
Therefore,
α + β + γ = -(Cofficient of x²)/(Cofficient of x³) = (-b)/a = 2/1
αβ + βγ + γα = (Cofficient of x)/(Cofficient of x³) = c/a = (-7)/1
αβγ = -(Constant term)/Cofficient of x³) = (-d)/a = (-14)/1
On comparing, a = 1, b = -2, c = -7 and d = 14
Hence, the required cubic polynomial is p(x) = x³ – 2x² – 7x + 14.

Here is the video explanation 😃

The zeroes of a polynomial are the values of the variable that make the polynomial equal to zero. The zeroes of a polynomial are also sometimes called the roots of the polynomial. Zeroes can also be used to factor polynomials, which can simplify equations and make them easier to solve.

Here is the Solution of this Question 😃👇