Question

Prove that:
(i) $\frac{cos(A+B+C)+cos(-A+B+C)+cos(A-B+C)+cos(A+B-C)}{sincos(A+B+C)+sincos(-A+B+C)+sincos(A-B+C)-sin\left(cos\right(A+B-C\left)\right)}$
(ii) $sin(B-C)cos(A-D)+sin(C-A)cos(B-D)+sin(A-B)cos(C-D)=0$

Answer 1

(i) We have
$LHS=\frac{cos(A+B+C)+cos(-A+B+C)+cos(A-B+C)+cos(A+B-C)}{sincos(A+B+C)+sincos(-A+B+C)+sincos(A-B+C)-sin\left(cos\right(A+B-C\left)\right)}2cos\left\{\frac{A+B+C-A+B+C}{2}\right\}cos\left\{\frac{A+B+C+A-B-C}{2}\right\}+2cos\left\{\frac{A-B+C+A+B-C}{2}\right\}=\frac{cos\frac{A-B+C-A-B+C}{2}}{2sin\frac{A+B+C-A-A+B+C}{2}cos\left\{\frac{A+B+C+A-B-C}{2}\right\}+2sin\left\{\frac{A-B+C-A+B+C}{2}\right\}cos\left\{\frac{A-B+C+A+B-C}{2}\right\}}=\frac{2cos(B+C)cosA+2cos(C-B)}{2sin(B+C)cosA+2sin(C-B)}=\frac{2cos\left\{\frac{B+C+C-B}{2}\right\}cos\left\{\frac{B+C-C+B}{2}\right\}}{2sin\left\{\frac{B+C+C-B}{2}\right\}cos\left\{\frac{B+C-C+B}{2}\right\}}=\frac{2cosCcosB}{2sinCcosB}=\frac{cosC}{sinC}=cotC=RHS\therefore \frac{cos(A+B+C)+cos(-A+B+C)+cos(A-B+C)+cos(A+B-C)}{sin(A+B+C)+sin(-A+B+C)+sin(A-B+C)-sin(A+B-C)}=cotC$

(ii) We have,
$LHS=sin(B-C)cos(A-D)+sin(C-A)cos(B-D)+sin(A-B)cos(C-D)=0=\frac{1}{2}\left[2sin\right(B-C\left)cos\right(A-D)+2sin(C-A\left)cos\right(B-D)+2sin(A-B\left)cos\right(C-D\left)\right]=\frac{1}{2}\left[sin\right(B-C+A-D)+sin(B-C-A+D)+sin(C-A+B-D)+sin(C-A-B+D)+sin(A-B+C-D)+sin(A-B-C+D\left)\right]=\frac{1}{2}\left[sin\right(A+B-C-D)+sin(B+D-C-A)+sin(B+C-A-D)+sin(C+D-A-B)+sin(A+C-B-D)+sin(A+D-B-C\left)\right]=\frac{1}{2}\left[sin\right(A+B-C-D)-sin(A+C-B-D)-sin(A+D-B-C)-sin(A+D-B-C)-sin(A+B-C-D)+sin(A+C-B-D)+sin(A+D-B-C\left)\right]=\frac{1}{2}\left[0\right]=RHS\therefore sin(B-C)cos(A-D)+sin(C-A)cos(B-D)+sin(A-B)cos(C-D)=0$