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Question Number 3536 by Syaka last updated on 14/Dec/15

prove  ((1 + cos x − cos y + cos z)/(1 + cos x + cos y − cos z)) = cot (1/2)z . tan (1/2)y

$${prove} \\ $$$$\frac{\mathrm{1}\:+\:{cos}\:{x}\:−\:{cos}\:{y}\:+\:{cos}\:{z}}{\mathrm{1}\:+\:{cos}\:{x}\:+\:{cos}\:{y}\:−\:{cos}\:{z}}\:=\:{cot}\:\frac{\mathrm{1}}{\mathrm{2}}{z}\:.\:{tan}\:\frac{\mathrm{1}}{\mathrm{2}}{y} \\ $$$$ \\ $$

Commented by Yozzii last updated on 14/Dec/15

x=y=0 , z=π/2  ∴ lhs=((1+1−1+0)/(1+1+1−0))=(1/3)  rhs=cot(π/4)tan(0/2)=1×0=0  lhs≠rhs.

$${x}={y}=\mathrm{0}\:,\:{z}=\pi/\mathrm{2} \\ $$$$\therefore\:{lhs}=\frac{\mathrm{1}+\mathrm{1}−\mathrm{1}+\mathrm{0}}{\mathrm{1}+\mathrm{1}+\mathrm{1}−\mathrm{0}}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${rhs}={cot}\frac{\pi}{\mathrm{4}}{tan}\frac{\mathrm{0}}{\mathrm{2}}=\mathrm{1}×\mathrm{0}=\mathrm{0} \\ $$$${lhs}\neq{rhs}. \\ $$

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