Question

Prove that:
(i) $\frac{sin(A+B)+sin(A-B)}{cos(A+B)+cos(A-B)}=tanA$
(ii) $\frac{sin(A-B)}{cosAcosB}+\frac{sin(B-C)}{cosBcosC}+\frac{sin(C-A)}{cosCcosA}=0$
(iii) $\frac{sin(A-B)}{sinAsinB}+\frac{sin(B-C)}{sinBsinC}+\frac{sin(C-A)}{sinCsinA}=0$
(iv) $si{n}^{2}B=si{n}^{2}A+si{n}^2(A-B)-2sinAcosBsin(A-B)$
(v) $co{s}^2+co{s}^{2}B-2cosAcosBcos(A+B)=si{n}^2(A+B)$
(vi) $\frac{tan(A+B)}{cot(A-B)}=\frac{ta{n}^{2}A-ta{n}^{2}B}{1-ta{n}^{2}Ata{n}^{2}B}$

Answer 1

(i) We have,
$LHS:\frac{sin(A+B)+sin(A-B)}{cos(A+B)+cos(A-B)}$
$=\frac{2sinAcosB}{2cosAcosB}$
$=\frac{sinA}{cosA}$
=tanA
=RHS
$\therefore LHS=RHS$
Hence proved.
(ii) We have,
$LHS:\frac{sin(A-B)}{cosAcosB}+\frac{sin(B-C)}{cosBcosC}+\frac{sin(C-A)}{cosCcosA}$
$=\frac{sinAcosB-cosAsinB}{cosAcosB}+\frac{sinBcosC-cosBsinC}{cosBcosC}+\frac{sinCcosA-cosCsinA}{cosCcosA}$
$=\frac{sinAcosB}{cosAcosB}-\frac{cosAsinB}{cosAcosB}+\frac{sinBcosC}{cosBcosC}-\frac{sinBsinC}{cosBcosC}+\frac{sinCcosA}{cosCcosA}-\frac{cosCsinA}{cosCcosA}$
=tanA-tanB+tanB-tanC+tanC-tanA
=0
=RHS
$\therefore LHS=RHS$
Hence proved.
(iii)We have,
LHS= $\frac{sin(A-B)}{sinAsinB}+\frac{sin(B-C)}{sinBsinC}+\frac{sin(C-A)}{sinCsinA}$
$\frac{sinAcosB-cosAsinB}{sinAsinB}+\frac{sinBsinC-cosBsinC}{sinBsinB}+\frac{sinCcosA-cosCsinA}{sinCsinA}$
$=\frac{sinAcosB}{sinAsinB}-\frac{cosAsinB}{sinAsinB}+\frac{sinBcosC}{sinBsinC}-\frac{cosBsinC}{sinBsinC}+\frac{sinCcosA}{sinCsinA}-\frac{cosCsinA}{sinCsinA}$
$=\frac{cosB}{sinB}-\frac{cosA}{sinA}+\frac{cosC}{sinC}-\frac{cosB}{sinB}+\frac{cosA}{sinA}-\frac{cosC}{sinC}$
=cotB-cotA+cotC-cotB+cotA-cotC
=0
=RHS
$\therefore LHS=RHS$
Hence proved.
(iv)We have,
$RHS:si{n}^{2}A+si{n}^2(A-B)-2sinAcosBsin(A-B)$
$=si{n}^{2}A=sin(A-B)\left[sin\right(A-B)-2sinAcosB]$
$=si{n}^{2}A+sin(A-B)[sinAcosB-cosAsinB-2sinAcosB]$
$=si{n}^{2}A+sin(A-B)[-sinAcosB-cosAsinB]$
$=si{n}^{2}A-sin(A-B)(sinAcosB+cosAsinB)$
$=si{n}^{2}A-sin(A-B)\left(sin\right(A+B\left)\right)$
$=si{n}^{2}A-sin(A-B)sin(A+B)$
$=si{n}^{2}A-(si{n}^{2}A-si{n}^{2}B)$
$=si{n}^{2}A-si{n}^{2}A+si{n}^{2}B$
$=si{n}^{2}B$
=LHS
$\therefore LHS=RHS$
Hence proved.
(v)RHS $=co{s}^{2}A+co{s}^{2}B-2cosAcosBcos(A+B)$
$=co{s}^{2}A+(1-si{n}^{2}B)-2cosAcosBcos(A+B)$
$=[co{s}^{2}A-si{n}^{2}B]-2cosAcosBcos(A+B)+1$
$=\left[cos\right(A+B\left)cos\right(A-B\left)\right]-2cosAcosBcos(A+B)+1$
$=cos(A+B)\left[cos\right(A-B)-2cosAcosB]+1$
$=cos(A+B)[cosAcosB-sinAsinB-2cosAcosB]+1$
$=cos(A+B)[-cosAcosB+sinAsinB]+1$
=-cos(A+B)[cos(A+B)]+1
=-$co{s}^2(A+B)+1$
=-$co{s}^{2}A+B$
= $1-co{s}^2(A+B)$
= $si{n}^2(A+B)$
=RHS
$\therefore LHS=RHS$
Hence proved.
(vi) We have,
LHS= $\frac{tan(A+B)}{cot(A-B)}$
$=\frac{tan(A+B)}{\frac{1}{tan(A-B)}}$
$=tan(A+B)tan(A-B)$
$=\left[\frac{tanA+tanB}{1-tanAtanB}\right]\left[\frac{tanA-tanB}{1+tanAtanB}\right]$
$\frac{(tanA+tanB)(tanA-tanB)}{(1-tanAtanB)(1+tanAtanB)}$
$=\frac{ta{n}^{2}A-ta{n}^{2}B}{1-(tanAtanB{)}^2}$
$=\frac{ta{n}^{2}A-ta{n}^{2}B}{1-ta{n}^{2}Ata{n}^{2}B}$
=RHS
$\therefore LHS=RHS$
Hence proved.