4.1 cos B 4.2.1 sin² P 4.2.2 sin β cos β 4.3 2sin(π + B)·cos(2π - B) - cos(180° - B) + cos(180° + B) = (−2 sin B)·cos B(− tan B) + (− cos(180° - B)) = 2sin B·cos B - cos B·2cos B = 2(sin² B + cos² B) = 2(1) = 2 4.4 cos θ + sin² θ·sec θ = tan θ cosec θ LHS / LK = cos θ / sin θ·sec θ / cosec θ = 1 / sin θ = cos² θ + sin² θ = 1 - NSC Technical Mathematics - Question 4 - 2023 - Paper 2

The expression gives the value of sin² P, indicating its identity.

Start with sin β/cos β and recognize that this can be simplified.

Evaluate the trigonometric identities involved in this expression.

1. Rewrite the equation using known values:

   $= (−2 sin B)·cos B(− tan B) + (− cos(180° - B))$

2. Continue to simplify: $= 2sin B·cos B - cos B·2cos B$

3. Factor the terms: $= 2(sin² B + cos² B)$

4. Utilize the Pythagorean identity: $= 2(1)$

5. Thus, the result is: $= 2$

Examine both sides of the equation:

1. Begin with: $LHS / LK = \frac{cos θ}{sin θ}\cdot\frac{sec θ}{cosec θ}$
2. Simplify: $= \frac{1}{sin θ}$
3. Recognize that: $= cos² θ + sin² θ$
4. Final expression: $= 1$