MHB Proving Trig Identity: (Sin2x-tanx)/cos2x=tanx

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The discussion focuses on proving the trigonometric identity (Sin2x - tanx) / cos2x = tanx. Participants suggest using the identities sin 2x = 2 sin x cos x and tan x = sin x / cos x to simplify the right-hand side. The next step involves combining the terms in the numerator into a single fraction with a common denominator. Factorization of the resulting expression is recommended to facilitate further simplification. The goal is to demonstrate the equality through these algebraic manipulations.
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May someone kindly assist me to prove this trig identity

(Sin2x-tanx) / cos2x = tanx

Thank you for your assistance
 
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paix1988 said:
May someone kindly assist me to prove this trig identity

(Sin2x-tanx) / cos2x = tanx

Thank you for your assistance

Making use of the identities $\sin 2x = 2 \sin x \cos x$ and $\tan x = \sin x / \cos x$ we can write the RHS as:
$$\frac{2 \sin x \cos x -\frac{\sin x}{\cos x}}{\cos 2x}.$$
To proceed, write the numerator as one fraction (thus with denominator $\cos x$) and factorize!
 
If there are an infinite number of natural numbers, and an infinite number of fractions in between any two natural numbers, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and an infinite number of fractions in between any two of those fractions, and... then that must mean that there are not only infinite infinities, but an infinite number of those infinities. and an infinite number of those...

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