InterviewSolution
Saved Bookmarks
| 1. |
Prove that:`tan6^0tan42^0tan66^0tan78^0=1.` |
|
Answer» `L.H.S. = tan6^@tan42^@tan66^@tan78^@` `=(sin6^@sin42^@sin66^@sin78^@)/(cos6^@cos42^@cos66^@cos78^@)` `=((2sin6^@sin66^@)(2sin42^@sin78^@))/((2cos6^@cos66^@)(2cos42^@cos78^@))` `=((cos60^@-cos72^@)(cos36^@-cos120^@))/((cos60^@+cos72^@)(cos36^@+cos120^@))` `=((1/2-sin18^@)(cos36^@+1/2))/((1/2+sin18^@)(cos36^@-1/2))...[As costheta = sin(90-theta)]` `=((1-2sin18^@)(2cos36^@+1))/((1+2sin18^@)(2cos36^@-1))` Now, putting, `sin18^@ = (sqrt5-1)/4 and cos36^@ = (sqrt5+1)/4` `=((1-2((sqrt5-1)/4))((2(sqrt5+1)/4)+1))/((1+2((sqrt5+1)/4))(2((sqrt5-1)/4)-1))` `=((3-sqrt5)(sqrt5+3))/((1+sqrt5)(sqrt5-1))` `=(9-5)/(5-1)` `=4/4 =1 = R.H.S.` |
|