In a triangle ABC- which is not right angled- then - cos A-cos B-cosec C -

∑ cos A cosec B cosec C =∑cos A/sin B sin C = ∑cos (B+c)/sin B sin c[In Δ ABC, A+B+C =π] = ∑cos B cos C sin B sin C/sin B sin C =∑ (1 cot A cot B) =∑ 1 ∑ ...

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Byju's Answer

Standard XI

Mathematics

Conditional Identities

In a triangle...

Question

In a triangle ABC, which is not right angled, then $\sum$ cos A.cos B.cosec C =___

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Solution

The correct option is B 2
$\sum c o s A c o s e c B c o s e c C = \sum \frac{c o s A}{s i n B s i n C} = - \sum \frac{c o s (B + c)}{s i n B s i n c} [I n Δ A B C, A + B + C = π] = - \sum \frac{c o s B c o s C - s i n B s i n C}{s i n B s i n C} = \sum (1 - c o t A c o t B) = \sum 1 - \sum c o t A c o t B$
=3-1
=2
{In $Δ A B C$ cot A cot B +cot B cot C +cot C cot A=}

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In a triangle ABC, which is not right angled, then ∑ cos A.cos B.cosec C =___

The correct option is B 2∑cosAcosecBcosecC=∑cosAsinBsinC=−∑cos(B+c)sinBsinc[InΔABC,A+B+C=π]=−∑cosBcosC−sinBsinCsinBsinC=∑(1−cotAcotB)=∑1−∑cotAcotB =3-1 =2 {In ΔABC cot A cot B +cot B cot C +cot C cot A=}

In a triangle ABC, which is not right angled, then $\sum$ cos A.cos B.cosec C =___