Answer:
[tex]\textsf{D)}\quad \sec^2\theta=1-\tan^2\theta[/tex]
Step-by-step explanation:
To determine which of the given equations is not a correction relation, we can use the Pythagorean Identity, sin²θ + cos²θ = 1.
[tex]\dotfill[/tex]
If we subtract cos²θ from both sides of the Pythagorean Identity, we get:
[tex]\sin^2\theta +\cos^2\theta-\cos^2\theta= 1- \cos^2\theta \\\\ \sin^2\theta = 1- \cos^2\theta[/tex]
Therefore, sin²θ = 1 - cos²θ is a correct relation.
[tex]\dotfill[/tex]
Divide both sides of the Pythagorean Identity by cos²θ:
[tex]\dfrac{\sin^2\theta}{\cos^2\theta} +\dfrac{\cos^2\theta}{\cos^2\theta} =\dfrac{1}{\cos^2\theta} \\\\\\ \tan^2\theta+1=\sec^2\theta[/tex]
Switch sides:
[tex]\sec^2\theta=1 + \tan^2\theta[/tex]
Therefore, sec²θ = 1 + tan²θ is a correct relation.
[tex]\dotfill[/tex]
Divide both sides of the Pythagorean Identity by sin²θ:
[tex]\dfrac{\sin^2\theta}{\sin^2\theta} +\dfrac{\cos^2\theta}{\sin^2\theta} =\dfrac{1}{\sin^2\theta} \\\\\\ 1+\cot^2\theta=\text{cosec}^2\theta[/tex]
Switch sides:
[tex]\text{cosec}^2\theta=1 + \cot^2\theta[/tex]
Therefore, cosec²θ = 1 + cot²θ is a correct relation.
[tex]\dotfill[/tex]
Divide both sides of the Pythagorean Identity by cos²θ:
[tex]\dfrac{\sin^2\theta}{\cos^2\theta} +\dfrac{\cos^2\theta}{\cos^2\theta} =\dfrac{1}{\cos^2\theta} \\\\\\ \tan^2\theta+1=\sec^2\theta[/tex]
Switch sides:
[tex]\sec^2\theta=1 + \tan^2\theta[/tex]
Therefore, sec²θ = 1 - tan²θ is NOT a correct relation.