急求助!!F.4 A.maths

1
tanA + tanB + tanC
=tan(180-(B+C)) + tanB + tanC
=-tan (B+C) + tanB + tanC
=[(tanB+tanC)/(tanBtanC-1)]+ tanB + tanC
=(tanB + tanC)[((tanBtanC-1)+1)/(tanBtanC-1)]
=tanBtanC(tanB + tanC)/(tanBtanC-1)
=tanBtanC[-tan(B+C)]
=tanBtanCtanA
(b)
tanA :tanB: tanC = 1:2:-6
tanB=2tanA; tanC =-6tanA
tanA + tanB + tanC =tanAtanBtanC
tanA + 2tanA -6 tanA =tanA(2tanA)(-6tanA)
-3tanA=-12tan^3A
tanA-4tan^3A=0
tanA(1-4tan^2A)=0
tanA=0 (rejected) or tanA=1/2 or tanA=-1/2 (rejected)
tanA=1/2
A=26.565
B=45
C=108.435
2 Given that x + ∮ = 兀/4, exprees tan ∮ in term of tanx and hence prove that ( 1 + tan x)(1+tan∮)=2.Deduce that tan兀/8 =厂2 -1
tan (x+∮)
=(tanx+tan∮)/(1-tanxtan∮)
(tanx+tan∮)/(1-tanxtan∮)=1
(tanx+tan∮)=(1-tanxtan∮)
tanx-1=-tan∮(tanx+1)
tan∮=(1-tanx)/(1+tanx)
( 1 + tan x)(1+tan∮)
=( 1 + tan x)[1+(1-tanx)/(1+tanx)]
=( 1 + tan x)[(1+tanx)+(1-tanx)/(1+tanx)]
=(1+tanx)+(1-tanx)
=2
( 1 + tan x)(1+tan∮)=2
1+(tanx+tan∮)+tanxtan∮=2
(tanx+tan∮)+tanxtan∮=1

if I let x=∮=兀/8
then
tan兀/8+tan兀/8+tan^2兀/8=1
tan^2兀/8+2tan兀/8-1=0
tan兀/8
=1/2[-2+厂(4+4)]
=1/2[-2+2厂2]
=厂2 -1
3
(a)
cos(x +a) = p and sin(x + B) = q.
cosxcosa-sinxsina=p
sinxcosB+cosxsinB=q
So
cosxcosacosB-sinxsinacosB=pcosB
sinxcosBsina+cosxsinBsina=qsina
add together that is
cosxcosacosB+cosxsinBsina=pcosB+qsina
cosx(cosacosB+sinBsina)=pcosB+qsina
cosxcos(a-B)=pcosB+qsina...(1)
On the other hand
cosxcosasinB-sinxsinasinB=psinB
sinxcosBcosa+cosxsinBcosa=qcosa
second minus first
sinxcosBcosa+sinxsinasinB=qcosa-psinB
sinx(cosBcosa+sinasinB)=qcosa-psinB
sinxcos(B-a)=qcosa-psinB...(2)
cosxcos(a-B)=pcosB+qsina...(1)
Now from (1)
cosxcos(B-a)=pcosB+qsina...(3)
sinx=[qcosa-psinB]/cos(B-a)
cosx=[pcosB+qsina]/cos(B-a)
(b)
sin^2x=[qcosa-psinB]^2/cos(B-a)^2
cos^2x=[pcosB+qsina]^2/cos(B-a)^2
[qcosa-psinB]^2+[pcosB+qsina]^2=cos^2(B-a)
q^2+p^2-2pqcosasinB+2sinacosB=cos^2(B-a)
p^2 + q^2 + 2pqsin(a-B) = cos^2(a - B).

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