# Prove that: (1+ tan 24°)(1+ tan 21°)(1+ tan 23°)(1+ tan 22°) = 4

sonika p
129 Points
6 years ago
[(1+ tan 24°)(1+ tan 21°)] [(1+ tan 23°)(1+ tan 22°) ][1+tan24+tan21+tan24tan21] [1+tan23+tan22+tan23tan22].......(1)tan(45)=1=tan(24+21) = tan(23+22)tan(A+B) = (tanA+tanB)/(1-tanAtanB)tan(24+21) = (tan24+tan21)/(1-tan24tan21)1*(1-tan24tan21)=tan24+tan211=tan24+tan21+tan24tan21....(2)similarly, tan23+tan22+tan23tan22=1......(3)substituting equation 2 & 3 in equation 1[1+1] [1+1]=4
32 Points
4 years ago
= [1+tan(45-21)].(1+tan21).[1+tan(45-22)].(1+tan22)
use the formula of tan(A+B)
= [1+  tan45-tan21     ].(1+tan21). [1+  tan45-tan22    ].(1+tan22)
1+tan45.tan21                             1+tan45.tan22
= [1+ 1-tan21  ] .(1+tan21).[1 + 1-tan22 ].(1+tan22)                             [tan45=1]
1+tan21                             1+tan22
lets take LCM,
= ( 1+tan21+1-tan21 ).(1+tan22).(1+tan22+1-tan22). (1+tan22)
1+tan21                                   1+tan22
cancelling the +ve & -ve term, and equal term of numerator and denominator.
= 2*2
= 4 = RHS
Proved