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in Trigonometry by (65.2k points)

If A + B + C = \(\frac{\pi}{2}\). P.T tanA · tanB + tanB . tanC + tanC · tanA = 1

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Given A + B + C = \(\frac{\pi}{2}\); A + B = 90° – C; tan(A + B) = tan(90° – C)

\(\frac{tanA + tanB}{1 - tanAtanB} = cotc \)

\(\frac{1}{tanC}\)

⇒ tanA tanC + tanB tanC = 1 – tanA · tanB 

⇒ tan A tanB + tanB tanC + tanC tanA = 1.

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