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Basic identities you can use:
Tip: When proving, convert everything to sines and cosines first, then simplify.
Start with \( \sin^2 x + \cos^2 x = 1 \).
Divide both sides by \( \cos^2 x \):
\( \dfrac{\sin^2 x}{\cos^2 x} + 1 = \dfrac{1}{\cos^2 x} \).
Replace \( \dfrac{\sin^2 x}{\cos^2 x} \) with \( \tan^2 x \):
\( \tan^2 x + 1 = \dfrac{1}{\cos^2 x} \).
Hence \( 1 + \tan^2 x = \dfrac{1}{\cos^2 x} \).
From Example 1 we know \( 1 + \tan^2 x = \dfrac{1}{\cos^2 x} \).
Take the reciprocal of both sides:
\( \dfrac{1}{1+\tan^2 x} = \cos^2 x \).
Thus \( \cos^2 x = \dfrac{1}{1+\tan^2 x} \).
This is just a direct substitution definition.
By definition \( \tan x = \dfrac{\sin x}{\cos x} \).
So \( \tan x = \dfrac{\sin x}{\cos x} \) holds true.
Subtract \( \cos^2 x \) from both sides:
\( \sin^2 x = 1 - \cos^2 x \).
Hence \( 1 - \cos^2 x = \sin^2 x \).
Square both sides:
\( \tan^2 x = \dfrac{\sin^2 x}{\cos^2 x} \).
Hence \( \dfrac{\sin^2 x}{\cos^2 x} = \tan^2 x \).
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