Q2=In an Acute angled Triangle ABC,Prove That secA(1+secA)secB(1+secB)secC(1+secC)is >=216
Asked by
| 28th Oct, 2012,
11:47: AM
Expert Answer:
Minimum value of the given expression will occur when triangle will be equilateral so A=B=C=?/3
Minimum value of expression will be=27*8=216
secA(1+secA)secB(1+secB)secC(1+secC)is >=216
Minimum value of the given expression will occur when triangle will be equilateral so A=B=C=?/3
Minimum value of expression will be=27*8=216
secA(1+secA)secB(1+secB)secC(1+secC)is >=216
Answered by
| 18th Nov, 2012,
07:51: PM
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