Let's solve the equation [tex]\(\sec(2A) = \frac{\cot^2(A) + 1}{\cot^2(A) - 1}\)[/tex] step by step.
First, let's rewrite some trigonometric identities for clarity:
- [tex]\(\sec(2A) = \frac{1}{\cos(2A)}\)[/tex]
- [tex]\(\cot(A) = \frac{\cos(A)}{\sin(A)}\)[/tex], so [tex]\(\cot^2(A) = \left(\frac{\cos(A)}{\sin(A)}\right)^2 = \frac{\cos^2(A)}{\sin^2(A)}\)[/tex]
Given [tex]\(\sec(2A)\)[/tex], we know:
[tex]\[\sec(2A) = \frac{1}{\cos(2A)}\][/tex]
Next, let's recall the double-angle identity for cosine:
[tex]\[\cos(2A) = \cos^2(A) - \sin^2(A)\][/tex]
We also know from the Pythagorean identity:
[tex]\[\cos^2(A) + \sin^2(A) = 1\][/tex]
Thus,
[tex]\[\cos(2A) = 1 - 2\sin^2(A)\][/tex]
or
[tex]\[\cos(2A) = 2\cos^2(A) - 1\][/tex]
Moving on to the given equation:
[tex]\[\sec(2A) = \frac{\cot^2(A) + 1}{\cot^2(A) - 1}\][/tex]
Substitute [tex]\(\cot^2(A) = \frac{\cos^2(A)}{\sin^2(A)}\)[/tex]:
[tex]\[\sec(2A) = \frac{\frac{\cos^2(A)}{\sin^2(A)} + 1}{\frac{\cos^2(A)}{\sin^2(A)} - 1}\][/tex]
Combining these into a single fraction:
[tex]\[\sec(2A) = \frac{\frac{\cos^2(A) + \sin^2(A)}{\sin^2(A)}}{\frac{\cos^2(A) - \sin^2(A)}{\sin^2(A)}}\][/tex]
Since [tex]\(\cos^2(A) + \sin^2(A) = 1\)[/tex]:
[tex]\[\sec(2A) = \frac{\frac{1}{\sin^2(A)}}{\frac{\cos(2A)}{\sin^2(A)}}\][/tex]
[tex]\[\sec(2A) = \frac{1}{\cos(2A)}\][/tex]
Thus, the original equation holds true with any angle [tex]\(A\)[/tex] that satisfies the trigonometric identities above. This confirms that the equation [tex]\(\sec(2A) = \frac{\cot^2(A) + 1}{\cot^2(A) - 1}\)[/tex] is consistent with the given Pythagorean and double-angle identities in trigonometry. Therefore, the result is indeed valid and consistent with known trigonometric identities.
