How to prove [1-⁡(tanx)^2⁡]/[1⁡+(⁡tanx)^2⁡] ⁡= [(cosx)^2⁡⁡]-[(⁡sinx)^2⁡]?

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Trigonometric Identities Part 2. Start solving the fractional form left-hand side members, [1-(tanx)^2]/[1+(tanx)^2], of the given equation until its right-hand side members, [(cosx)^2]-[(sinx)^2], are obtained. Both the numerator and the denominator has (tanx)^2 terms, replace it by [(sinx)^2]/[(cosx)^2]. The result is a complex fraction with [(sinx)^2] and [(cosx)^2] terms. Evaluate and simplify it.

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