Answer:
See Below.
Step-by-step explanation:
We want to verify the equation:
[tex]\displaystyle \frac{1}{1+\sin\theta} = \sec^2\theta - \sec\theta \tan\theta[/tex]
To start, we can multiply the fraction by (1 - sin(θ)). This yields:
[tex]\displaystyle \frac{1}{1+\sin\theta}\left(\frac{1-\sin\theta}{1-\sin\theta}\right) = \sec^2\theta - \sec\theta \tan\theta[/tex]
Simplify. The denominator uses the difference of two squares pattern:
[tex]\displaystyle \frac{1-\sin\theta}{\underbrace{1-\sin^2\theta}_{(a+b)(a-b)=a^2-b^2}} = \sec^2\theta - \sec\theta \tan\theta[/tex]
Recall that sin²(θ) + cos²(θ) = 1. Hence, cos²(θ) = 1 - sin²(θ). Substitute:
[tex]\displaystyle \displaystyle \frac{1-\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta \tan\theta[/tex]
Split into two separate fractions:
[tex]\displaystyle \frac{1}{\cos^2\theta} -\frac{\sin\theta}{\cos^2\theta} = \sec^2\theta - \sec\theta\tan\theta[/tex]
Rewrite the two fractions:
[tex]\displaystyle \left(\frac{1}{\cos\theta}\right)^2-\frac{\sin\theta}{\cos\theta}\cdot \frac{1}{\cos\theta}=\sec^2\theta - \sec\theta \tan\theta[/tex]
By definition, 1 / cos(θ) = sec(θ) and sin(θ)/cos(θ) = tan(θ). Hence:
[tex]\displaystyle \sec^2\theta - \sec\theta\tan\theta \stackrel{\checkmark}{=} \sec^2\theta - \sec\theta\tan\theta[/tex]
Hence verified.
