4.1 Given: tan(π + A) ⋅ cos(180° − A) ⋅ sin(360° − A) divided by sin(2π + A) 4.1.1 Simplify by reduction: tan(π + A) 4.1.2 Simplify: tan(π + A) ⋅ cos(180° − A) ⋅ sin(360° − A) divided by sin(2π + A) 4.2 Complete the identity: cot² x − cosec² x = 4.3 Prove the identity: sin x + cos² x − cosec x = cosec x - NSC Technical Mathematics - Question 4 - 2024 - Paper 2

First, simplify each component:

tan(π + A) = tan(A)

cos(180° - A) = -cos(A)

sin(360° - A) = -sin(A)

sin(2π + A) = sin(A)

Now, substitute these values into the original expression:

rac{ an(A) imes (- ext{cos}(A)) imes (- ext{sin}(A))}{ ext{sin}(A)}

This simplifies to:

tan(A) imes ext{cos}(A)

Therefore, the final answer is:

tan(A) imes ext{cos}(A)

Using the Pythagorean identity:

cot² x + 1 = cosec² x,

rearranging gives us:

cot² x = cosec² x − 1.

Thus, the completed identity is:

cot² x − cosec² x = −1

Begin by rewriting the left-hand side:

LHS:

sin x + cos² x − rac{1}{ ext{sin} x}

Using the identity: cos² x = 1 − sin² x,

we have:

LHS = sin x + (1 − sin² x) − rac{1}{ ext{sin} x} = 1 − sin x − rac{1}{ ext{sin} x}.

Now simplifying this:

rac{ ext{sin}^2 x}{ ext{sin} x} + 1 − rac{1}{ ext{sin} x} = rac{ ext{sin}^2 x + ext{sin} x − 1}{ ext{sin} x}.

The right-hand side:

RHS: rac{1}{ ext{sin} x} = cosec x.

Thus, the equation is equal to:

LHS = RHS, proving the identity: