cos(a+b)/cos(a-b)=sin(c+d)sin(c-d)
abhisheksp:
either u have not given full question content
i can answer bro
as i have answer it before in my referrence books
Answers
Answered by
0
HeY!!
cos(A+B)/cos(A-B)=sin(C+D)/sin(C-D)using P(x)=q(x)g(x)+r(x) process,
[cos(A-B)-cos(A+B)]/[cos(A-B)+cos(A+B)]
=[sin(C-D)-sin(C+D)]/[sin(C-D)+sin(C+D)] =2sinAsinB/[cos(A+B)+cos(A-B)]
=-[sin(C+D)-sin(C-D)]/[sin(C+D)+sin(C-D)]
=2sinAsinB/2cosAcosB
=2cosCsinD/2sinCcosD
=tanAtanB=-cotCtanD
= tanAtanB/cotC=-tanD= tanAtanBtanC=-tanD
Thus, tanAtanBtanC+tanD=0
Hence proved.
hope it helped u
mrk as brainliest
cos(A+B)/cos(A-B)=sin(C+D)/sin(C-D)using P(x)=q(x)g(x)+r(x) process,
[cos(A-B)-cos(A+B)]/[cos(A-B)+cos(A+B)]
=[sin(C-D)-sin(C+D)]/[sin(C-D)+sin(C+D)] =2sinAsinB/[cos(A+B)+cos(A-B)]
=-[sin(C+D)-sin(C-D)]/[sin(C+D)+sin(C-D)]
=2sinAsinB/2cosAcosB
=2cosCsinD/2sinCcosD
=tanAtanB=-cotCtanD
= tanAtanB/cotC=-tanD= tanAtanBtanC=-tanD
Thus, tanAtanBtanC+tanD=0
Hence proved.
hope it helped u
mrk as brainliest
Similar questions