Question

1.(1+cos A)(1+cos B)(1+cos C)=(1-cos A)(1-cos B)(1-cos C)
Show that one of the values of each member of this equality is sin A sin B
sin C. (HOTS)

Answer 1

This would be helpful to understand

Answer 2

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 .

•••••