How do I establish this identity??
2007-11-05 03:39:22 · 4 answers · asked by melissa_quintana1516 1 in Science & Mathematics ➔ Mathematics
tan (105)°
tan (60 + 45)°
(tan 60° + tan 45°) / (1 - tan 60° tan 45°)
(√3 + 1) / [ 1 - (√3/2)(1) ]
(√3 + 1) / [1 - √3/2 ]
2007-11-05 06:10:37 · answer #1 · answered by Como 7 · 2⤊ 0⤋
Tan !05 degree
= Tan (90+ 15) = - Cot 15 = - 1 / (Tan 15) = - 1 / Tan (60- 45)
= [1 + (Tan 60)(tan 45)] / Tan 60 - tan 45
= ( 1 + sq root 3) (1 + sqroot 3) / ( 3 - 1 )
=( 3 + 1 + 2. sq root 3) / 2 = 2 + sq root of 3 .......... Ans
2007-11-05 03:58:19 · answer #2 · answered by Pramod Kumar 7 · 0⤊ 0⤋
using the addition formulae
tan(a+b) = (tana+tanb) / (1 - (tana)(tanb))
using 45 degrees and 60 degrees would work
2007-11-05 03:43:34 · answer #3 · answered by Anonymous · 0⤊ 0⤋
tan(105 degree)
= tan(60degree+45degree)
= [tan(60deg)+tan(45deg)]
/[1-tan(60deg).tan(45deg)]
using tan(A+B)
=[tanA+tanB]/[1-tanA. tanB]
= [sqrt(3)+1]/[1-sqrt(3)]
using tan(60deg) =sqrt(3) &an(45deg)=1
=[sqrt(3)+1]/[1-sqrt(3)]*[1+sqrt(3)]/[1+sqrt(3)]
=(1+3+2sqrt(3))/(-2)=(4+2sqrt(3))/(-2)
=-[2+sqrt(3)]= -3.732 using sqrt(3) =1.732
2007-11-05 04:08:54 · answer #4 · answered by Dr K.L.Verma 2 · 0⤊ 0⤋