a maths請寫step..**急

(a)
tanAtan(A+B) = k
(sinA/cosA)[sin(A+B)/cos(A+B)]=k
sinAsin(A+B)=kcosAcos(A+B)
-1/2[cos(2A+B)-cos(-B)]=k/2[cos(2A+B)+cos(-B)]
-[cos(2A+B)-cosB]=k[cos(2A+B)+cosB]
-cos(2A+B)+cosB=kcos(2A+B)+kcosB
(k+1)cos(2A + B) = (1-k)cosB...(*)
(b)
sub B=60,k=2 in (*)
(2+1)cos(2A + 60) = (1-2)cos60
3cos(2A + 60) = -1/2
cos(2A + 60) = -1/6
2A + 60=360n+99.594 or 360n-99.594
2A=360n+39.594 or 360n-159.594
A= 180n+19.797 or 180n-79.797

a maths請寫step..**急
(a) Prove that if tanAtan(A+B) = k , then (k+1)cos(2A + B) = (1-k)cosB

(b) Find the general solution of tanAtan(A + 60° ) = 2

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(a)

tanAtan(A+B) = k

(sinA/cosA)[sin(A+B)/cos(A+B)]=k

sinAsin(A+B)=kcosAcos(A+B)

-1/2[cos(2A+B)-cos(-B)]=k/2[cos(2A+B)+cos(-B)]

-[cos(2A+B)-cosB]=k[cos(2A+B)+cosB]

-cos(2A+B)+cosB=kcos(2A+B)+kcosB

(k+1)cos(2A + B) = (1-k)cosB...(*)

(b)

sub B=60,k=2 in (*)

(2+1)cos(2A + 60) = (1-2)cos60

3cos(2A + 60) = -1/2

cos(2A + 60) = -1/6

2A + 60=360n+99.594 or 360n-99.594

2A=360n+39.594 or 360n-159.594

A= 180n+19.797 or 180n-79.797