1
Question
In a triangle ABC, which is not right angled, then ∑ cos A.cos B.cosec C =___
In a triangle ABC, which is not right angled, then ∑ cos A.cos B.cosec C =___
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Solution
The correct option is B 2
∑cosA cosec Bcosec C=∑cosAsinB sinC=−∑cos(B+C)sin B sin C [InΔABC, A+B+C=π]=−∑cos B cos C−sin B sin Csin B sin C=∑(1−cot A cot B)=∑1−∑cot A cot B
=3-1
=2
[In ΔABC, cot A cot B +cot B cot C +cot C cot A= 1]
2
∑cosA cosec Bcosec C=∑cosAsinB sinC=−∑cos(B+C)sin B sin C [InΔABC, A+B+C=π]=−∑cos B cos C−sin B sin Csin B sin C=∑(1−cot A cot B)=∑1−∑cot A cot B
=3-1
=2
[In ΔABC, cot A cot B +cot B cot C +cot C cot A= 1]
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