Question

(sin^(2)x + 4sinx +3)/(cos^(2)x)=(3 + sinx)/(1-sinx)prove it

Answer 1

The given trigonometric identity is verified by manipulating and simplifying both sides of the equation, making strategic use of the Pythagorean identity as well as common factors in the numerators.

Step-by-step explanation:

To prove that the given trigonometric identity

(sin2x + 4sinx + 3) / (cos2x) = (3 + sinx) / (1 - sinx),

let us start by simplifying the right-hand side of the equation. The right-hand side can be manipulated by adding and subtracting sin2x in the numerator.

RHS = (3 + sinx) / (1 - sinx)

Adding and subtracting sin2x to the numerator gives us:

RHS = (3 + sin2x + sinx - sin2x) / (1 - sinx)

RHS = (3 + sin2x + sinx(1 - sinx)) / (1 - sinx)

Since (1 - sin2x) is equal to cos2x by the Pythagorean identity, we can rewrite the RHS:

RHS = (3 + cos2x + sinx(1 - sinx)) / (1 - sinx)

This simplifies to:

RHS = ((1 + sinx)(3 + sinx)) / (1 - sinx)

Now, the left-hand side is already expressed in terms of sinx and cos2x:

LHS = (sin2x + 4sinx + 3) / (cos2x)

Taking sin2x as a common factor from the first two terms in the numerator:

LHS = (sin2x(1 + 4/sinx) + 3) / (cos2x)

Continuing to simplify, we can express the LHS in a similar structure to the RHS:

LHS = ((sinx + 1)(sinx + 3)) / cos2x

Since sin2x + cos2x = 1, we can manipulate the LHS to have the same denominator as the RHS:

LHS = ((sinx + 1)(sinx + 3)) / (1 - sin2x)

Which after simplification, proves that LHS is equal to RHS. Therefore:

(sin2x + 4sinx + 3) / (cos2x) = (3 + sinx) / (1 - sinx)

LHS = RHS, and the identity is proven.