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1.

If `A+B+C=pi`, prove that : `sin (B+C-A) + sin (C+A-B) + sin (A+B-C)=4sinA sinB sinC`

Answer» `LHS=sin(A+B+C-2A)+sin(A+B+C-2B)-sin(A+B+C-2C)`
`=sin (pi-2A)+sin(pi-2B)-sin(pi-2C)`
`=sin2A+sin2B-sin2C.`
Discussion
2.

If `A+B+C=pi`, prove that : `cos2A+cos2B-cos2C=1-4sinA sinB cosC`

Answer» We have
`LHS=(cos2A+cos 2B)-cos 2C`
`=2cos (A+B)cos (A-B)-2cos^(2)C+1`
`=2cos (pi-C)cos (A-B)-2cos^(2)C+1`
`=-2cos C cos (A-B)-2cos^(2)C+1`
`=1-2 cos C [ cos (A-B)+cosC]`
`=1-2cos C [cos(A-B)+cos{pi-(A+b)}]`
`=1-2cos C [ cos (A-B)-cos (A+B)]`
`=1-2 cos C [2sin A sin C=RHS.`
`therefore cos 2A+cos2B-cos2C=1-4sin A sin B cos C.`
Discussion
3.

If `A+B+C=pi`, prove that : `tan A/2 tan B/2 + tan B/2 tan C/2 + tan C/2 tan A/2 =1`

Answer» Here , `A+B+C=pi`
`implies((A)/(2)+(B)/(2))=((pi)/(2)-( C)/(2))`
`implies tan ((A)/(2)+(B)/(2))=tan ((pi)/(2)-( C)/(2))=cot ""(C)/(2)`
`implies (tan ""(A)/(2)+tan ""(B)/(2))/(1-tan""(A)/(2)tan""(B)/(2))=(1)/(tan""( C)/(2))`
`implies tan ""(C)/(2)tan ""(A)/(2)+tan ""(B)/(2)tan ""( C)/(2)=1-tan""(A)/(2) tan""(B)/(2)`
`implies tan""(A)/(2)tan ""(B)/(2)+tan ""(B)/(2)tan ""( C)/(2)+tan""(C )/(2)tan""(A)/(2)+1.`
Discussion
4.

If `A+B+C=pi` then prove that `cos^2 (A/2)+cos^2 (B/2)-cos^2 (C/2)=2cos(A/2)cos(B/2)sin(C/2)`

Answer» We have
`LHS=cos^(2)""(A)/(2)+cos^(2)""(B)/(2)-cos^(2)""(C)/(2)`
`=(1)/(2)(1+cosA)+(1)/(2)(1+cosB)-(1)/(2)(1+cos C)`
`=(1)/(2)+(1)/(2)(cosA+cos B-cos C)`
`=(1)/(2)+(1)/(2)[(4cos""(A)/(2)cos""(B)/(2)sin""(C )/(2))-1]`
`=2 cos ""(A)/(2)cos""(B)/(2)""sin""(C)/(2)=2cos ""(A)/(2)cos""(B)/(2)sin""( C)/(2).`
Discussion
5.

If `A+B+C=pi `, prove that `cos 2A +cos 2B +cos 2C=-1-4cos A cos Bcos C.`

Answer» We have
`LHS=cos 2A+cos 2B+cos 2C`
`=2cos(A+B)cos (A-B)+2cos^(2)C-1`
`=2cos (pi-C)cos (A-B)+2cos^(2)C-1`
`=-2cos C cos (A-B)+2cos^(2)C-1`
`=-2cos C[cos (A-B)-cosC]-1`
`=-1-2cos C[cos (A-B)-cos {pi-(A-B)}]`
`=-1-2cos C[cos(A-B)+cos(A+B) ]`
`=-1-4cos A cos B cos C =RHS.`
`therefore cos 2A +cos 2B +cos2C=-1-4cos A cos B cos C.`
Discussion