4.1 Simplify: 4.1.1 cot² 2β - cosec² 2β 4.1.2 tan² A - cosec² A - cos 2π 4.2 Given the identity: cosec(180° + θ) · sin(360° - θ) - [sin(180° + θ)] sec 60° = cos θ 4.2.1 Write down the numerical value of sec 60° - NSC Technical Mathematics - Question 4 - 2019 - Paper 2

We start with the equation:

$an^2 A - ext{cosec}^2 A - ext{cos} 2 heta$

Using the identity, $ext{cos} 2 heta = 1 - 2 ext{sin}^2 heta$ is relevant here:

an^2 A = rac{ ext{sin}^2 A}{ ext{cos}^2 A} \ ext{cosec}^2 A = rac{1}{ ext{sin}^2 A}

Substituting these values lets us simplify to:

rac{ ext{sin}^2 A}{ ext{cos}^2 A} - rac{1}{ ext{sin}^2 A} - 1

Ultimately, this simplifies to:

$an^2 A$

The numerical value of sec 60° can be derived from:

ext{sec} 60° = rac{1}{ ext{cos} 60°}

Since ext{cos} 60° = rac{1}{2}, we have:

ext{sec} 60° = rac{1}{rac{1}{2}} = 2

To prove the identity:

Starting with:

$ext{L.H.S} = ext{cosec}(180° + θ) · ext{sin}(360° - θ) - [ ext{sin}(180° + θ)] ext{sec} 60°$

Substituting known values, we note that:

• $ext{cosec}(180° + θ) = - ext{cosec} θ$
• $ext{sin}(360° - θ) = - ext{sin} θ$
• $ext{sec} 60° = 2$

So, substituting these gives:

$- ext{cosec} θ imes (- ext{sin} θ) - (- ext{sin} θ) * 2$

This simplifies to:

$ext{cosec} θ ext{sin} θ + 2 ext{sin} θ = 1$

Thus, we arrive at:

$1 - ext{sin}^2 θ = ext{cos} θ = ext{R.H.S.}$