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Prove that: tan 36° + tan 9° + tan 36° tan 9° = 1

Answer»

We have 36° + 9° = 45° 

⇒ tan(36° + 9°) = tan 45°

We know that tan(A + B) = \(\frac{tanA+tanB}{1-tanAtanB}\)

⇒ \(\frac{tan36°+tan9°}{1-tan36°tan9°}\) = 1

⇒ tan 36° + tan 9° = 1 – tan 36° tan 9° 

∴ tan 36° + tan 9° + tan 36° tan 9° = 1 

Hence proved.



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