Question

(1+cos a)(1+cos b)(1+cos c)=(1-cos a)(1-cos b)(1-cos)=k find the value of k

Answer 1

Solution:

We have ,

(1+cos A)(1+cos B)(1+cos C)=(1-cos A)(1-cos B)(1-cos C)

Multiply both sides of the equation by (1-cosA)((1-cosB)(1-cosC) , we get

=> [(1+cosA)(1-cosA)(1+cosB)(1-cosB).

(1+cosC)(1-cosC)]=[(1-cosA)²(1-.

cosB)²(1-cosC)²]

=> [(1-cos²A)(1-cos²B)(1-cos²C)]=[(1-cosA).

(1-cosB)(1-cosC)]²

=> sin²Asin²Bsin²C=[(1-cosA)(1-cosB)(1-.

cosC)]²

=>[ sinAsinBsinC]²=[(1-cosA)(1-cosB)(1-.

cosC)]²

=> sinAsinBsinC=(1-cosA)(1-cosB)(1-cosC)

Hence , proved .

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