Question

tan18+tan27+tan18 tan27 =0

Answer 1

tan(A+B)=tanA+tanB/1-tanAtanB
tanA+tanB= tan(A+B)*(1-tanAtanB)
tan18+tan27=tan(18+27)*(1-tan18*tan27)
tan18+tan27=tan45*(1-tan18*tan27)
tan18+tan27 =1-tan18*tan27
tan18+tan27+tan18*tan27=1

Answer 2

Answer:

Ans:1

Step-by-step explanation:

tan18°+tan27°+tan18°tan27°=0

tan(18+27)=

$\frac{ \tan(18) + \tan(27) }{1 - \tan(18) \tan(27) }$

$\tan(45) = \frac{ \tan(18) + \tan(27) }{1 - \tan(18) \tan(27) }$

$1 - \tan(18) \tan(27) = \tan(18) + \tan(27)$

$1 = \tan(18 + 27) + \tan(18) \tan(27)$

$= 1$