Given that,
X=sin(θ+7π12)+sin(θ−π12)+sin(θ+3π12)
Y=cos(θ+7π12)+cos(θ−π12)+cos(θ+3π12)
Then,
X=sin(θ+7π12)+sin(θ−π12)+sin(θ+3π12)
=2sin(θ+7π12+θ−π12)2cos(θ+7π12−θ+π12)2+sin(θ+3π12)
=2sin(2θ+6π12)2cos8π24+sin(θ+3π12)
=2sin(θ+π4)cosπ3+sin(θ+3π12)
=2×12sin(θ+π4)+sin(θ+3π12)
=sin(θ+π4)+sin(θ+3π12)
X=2sin(θ+π4)
Similarly,
Y=cos(θ+7π12)+cos(θ−π12)+cos(θ+3π12)
=2cos(θ+7π12+θ−π12)2cos(θ+7π12−θ+π12)2+cos(θ+3π12)
=2cos(θ+π4)cosπ3+cos(θ+3π12)
=cos(θ+π4)+cos(θ+π4)
Y=2cos(θ+π4)
Put the value of X&Y in LHS we get,
LHS
XY−YX
=2sin(θ+π4)2cos(θ+π4)−2cos(θ+π4)2sin(θ+π4)
=sin(θ+π4)cos(θ+π4)−cos(θ+π4)sin(θ+π4)
=tan(θ+π4)−cot(θ+π4)
=tanθ+tanπ41−tanθtanπ4−coθcotπ4−1coθ+cotπ4
=1+tanθ1−tanθ−cotθ−1cotθ+1
=1+tanθ1−tanθ−1tanθ−11tanθ+1
=1+tanθ1−tanθ−1−tanθ1+tanθ
=(1+tanθ)2−(1−tan2θ)1−tan2θ
=1+tan2θ+2tanθ−1−tan2θ+2tanθ1−tan2θ
=4tanθ1−tan2θ
=2×2tanθ1−tan2θ
=2tan2θ
=RHS