If `x^5 = (4 - 3i)^5,` then the product of all of its roots (where `theta = -tan^(-1)(3/4)`)

If x^5=(4-3i)^5 thrn the product of all of its roots (where theta=-tan^(-1)(3/4)

The continued product of all values of root(5)(4) is

Solve x^4-5x^3+5x^2+5x-6 =0 given that the product of two of its roots is 3

The angle between the normals of ellipse 4x^2 + y^2 = 5 , at the intersection of 2x+y=3 and the ellipse is (A) tan^(-1) (3/5) (B) tan^(-1) (3/4) (C) tan^(-1) (4/3) (D) tan^(-1) (4/5)

The angle between the normals of ellipse 4x^2 + y^2 = 5 , at the intersection of 2x+y=3 and the ellipse is (A) tan^(-1) (3/5) (B) tan^(-1) (3/4) (C) tan^(-1) (4/3) (D) tan^(-1) (4/5)

If tan 2theta . Tan 3 theta =1 , then sin5theta =

If 4 sin theta + 3 cos theta =5 then tan theta =

If 5tan theta-4=0, then the value of (5sin theta-4cos theta)/(5sin theta+4cos theta) is (5)/(3)(b)(5)/(6)(c)0(d)(1)/(6)

If 5 tan theta = 4, then the value of (5 sin theta - 3 cos theta)/(5 sin theta + 3 cos theta) is