COS(A-B)/COS(A+B)+COS(C+D)/COS(C-D)=0 THEN SHOW THAT tan A tan B tan C tan D=-1

siddharth gupta
28 Points
9 years ago
STEP1:COS(A-B)/COS(A+B)=(COS(A)COS(B)+SIN(A)SIN(B))/(COS(A)COS(B)-SIN(A)SIN(B))
SIMILARLY OPEN THE EXPRESSION FOR THE OTHER TERM.
STEP2:DIVIDE NUMERATOR AND DENOMINATOR IN BOTH CASES BY COS(A)COS(B) AND COS(C)COS(D) RESPECTIVELY.
STEP3:SOLVE THE EQUATION FORMED IN TAN TO OBTAIN YOUR RESULT.
Ashish
20 Points
9 years ago
we get tanAtanBtanCtanD=1 not -1

Ashish
20 Points
9 years ago
we get tanAtanBtanCtanD=1 not -1

Ashish
20 Points
9 years ago
we get tanAtanBtanCtanD=1 not -1

Rishav
13 Points
7 years ago
Use the formula of cos (A-B)/co(A+B) then similarly use on other side ,then divide numerator and denominator by cos(A)cos(B) and after U will get tan A.tanB.tanC.tanD=1
Soumendu Majumdar
159 Points
6 years ago
{ cos(A-B)/cos(A+B)} + { cos(C+D)/cos(C-D)}=0
Break the above like cos(A+B)=cosAcosB-sinAsinB & cos(A-B)=cosAcosB+sinAsinB
then divide numerator and denominator by cosAcosB & cosCcosD respectively.
therefore we get: {(1+tanAtanB)/(1-tanAtanB)} + {(1-tanCtanD)/(1+tanCtanD)}=0
Simplifying it you’ll get
2tanAtanBtanCtanD=-2
implies: tanAtanBtanCtanD=-1 [Proved]

Hope it helps...