Q.

The least value of secA+secB+secC in a acute angled triangle

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a

−18

b

332

c

−12

d

6

answer is D.

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Detailed Solution

In ΔABC ,A,B,C are acute. ⇒secA,secB,secC are positive  Now, A·M≥H·M.secA+secB+secC3≥3cosA+cosB+cosC                             →1We have cosA+cosB+cosC3≤cosA+B+C3                                                  ≤cos60°∴  cosA+cosB+cosC≤32  →2From (1) and (2) secA+secB+secC≥6
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