Question

# Prove that: (i) sin(A+B)+sin(A−B)cos(A+B)+cos(A−B)=tanA (ii) sin(A−B)cosAcosB+sin(B−C)cosBcosC+sin(C−A)cosCcosA=0 (iii) sin(A−B)sinAsinB+sin(B−C)sinBsinC+sin(C−A)sinCsinA=0 (iv) sin2B=sin2A+sin2(A−B)−2sinAcosBsin(A−B) (v) cos2+cos2B−2cosAcosBcos(A+B)=sin2(A+B) (vi) tan(A+B)cot(A−B)=tan2A−tan2B1−tan2Atan2B

Solution

## (i) We have, LHS:sin(A+B)+sin(A−B)cos(A+B)+cos(A−B) =2sinAcosB2cosAcosB =sinAcosA =tanA =RHS ∴LHS=RHS Hence proved. (ii) We have, LHS:sin(A−B)cosAcosB+sin(B−C)cosBcosC+sin(C−A)cosCcosA =sinAcosB−cosAsinBcosAcosB+sinBcosC−cosBsinCcosBcosC+sinCcosA−cosCsinAcosCcosA =sinAcosBcosAcosB−cosAsinBcosAcosB+sinBcosCcosBcosC−sinBsinCcosBcosC+sinCcosAcosCcosA−cosCsinAcosCcosA =tanA-tanB+tanB-tanC+tanC-tanA =0 =RHS ∴LHS=RHS Hence proved. (iii)We have, LHS= sin(A−B)sinAsinB+sin(B−C)sinBsinC+sin(C−A)sinCsinA sinAcosB−cosAsinBsinAsinB+sinBsinC−cosBsinCsinBsinB+sinCcosA−cosCsinAsinCsinA =sinAcosBsinAsinB−cosAsinBsinAsinB+sinBcosCsinBsinC−cosBsinCsinBsinC+sinCcosAsinCsinA−cosCsinAsinCsinA =cosBsinB−cosAsinA+cosCsinC−cosBsinB+cosAsinA−cosCsinC =cotB-cotA+cotC-cotB+cotA-cotC =0 =RHS ∴LHS=RHS Hence proved.  (iv)We have, RHS:sin2A+sin2(A−B)−2sinAcosBsin(A−B) =sin2A=sin(A−B)[sin(A−B)−2sinAcosB] =sin2A+sin(A−B)[sinAcosB−cosAsinB−2sinAcosB] =sin2A+sin(A−B)[−sinAcosB−cosAsinB] =sin2A−sin(A−B)(sinAcosB+cosAsinB) =sin2A−sin(A−B)(sin(A+B)) =sin2A−sin(A−B)sin(A+B) =sin2A−(sin2A−sin2B) =sin2A−sin2A+sin2B =sin2B =LHS ∴LHS=RHS Hence proved. (v)RHS =cos2A+cos2B−2cosAcosBcos(A+B) =cos2A+(1−sin2B)−2cosAcosBcos(A+B) =[cos2A−sin2B]−2cosAcosBcos(A+B)+1 =[cos(A+B)cos(A−B)]−2cosAcosBcos(A+B)+1 =cos(A+B)[cos(A−B)−2cosAcosB]+1 =cos(A+B)[cosAcosB−sinAsinB−2cosAcosB]+1 =cos(A+B)[−cosAcosB+sinAsinB]+1 =-cos(A+B)[cos(A+B)]+1 =-cos2(A+B)+1 =-cos2A+B = 1−cos2(A+B) = sin2(A+B) =RHS ∴LHS=RHS Hence proved. (vi) We have, LHS= tan(A+B)cot(A−B) =tan(A+B)1tan(A−B) =tan(A+B)tan(A−B) =[tanA+tanB1−tanAtanB][tanA−tanB1+tanAtanB] (tanA+tanB)(tanA−tanB)(1−tanAtanB)(1+tanAtanB) =tan2A−tan2B1−(tanAtanB)2 =tan2A−tan2B1−tan2Atan2B =RHS ∴LHS=RHS Hence proved.MathematicsRD SharmaStandard XI

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