Question

Prove that tan 72° = tan 18° + tan 54°

Answer 1

we have to prove that tan72° = tan18° + 2tan54° [ you did mistake in typing ]

we know, tan(A - B) = (tanA - tanB)/(1 + tanA. tanB)

tan54° = tan(72° - 18°)

or, tan54° = (tan72° - tan18°)/(1 + tan72°. tan18°)

or, tan54°(1 + tan72°. tan18°) = tan72° - tan18°

we know, tan72° = tan(90° - 18°) = cot18°

or, tan54°(1 + cot18°. tan18°) = tan72° - tan18°

or, tan54° (1 + 1 ) = tan72° - tan18°

or, 2tan54° = tan72° - tan18°

hence, tan72° = tan18° + 2tan54°

LHS = RHS

Answer 2

HELLO DEAR,


IT SEEMS THERE IS SOME TYPING MISTAKE
YOUR QUESTIONS SHOULD BE tan72° = tan18° + 2tan54°


we know:-
tan(A - B) = (tanA - tanB)/(1 + tanA. tanB)
tan72° = tan(90° - 18°) = cot18°

tan54° = tan(72° - 18°)

=> tan54° = (tan72° - tan18°)/(1 + tan72°. tan18°)

=> tan54°(1 + tan72°. tan18°) = tan72° - tan18°

=> tan54°(1 + cot18°. tan18°) = tan72° - tan18°

=> tan54° (1 + 1 ) = tan72° - tan18°

=> 2tan54° = tan72° - tan18°

hence, tan72° = tan18° + 2tan54°



I HOPE IT'S HELP YOU DEAR,
THANKS