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Question Number 37218 by Rio Mike last updated on 10/Jun/18
Proove that a) ((1−cosA + cos B − cos(A+B))/(1+cos A − cosB−cos(A+B)))= tan(A/2).cot (B/2) b) cosα cos(60−α)cos(60+α)= (1/4)cos3α
$$\:{Proove}\:{that} \\ $$$$\left.{a}\right)\:\:\frac{\mathrm{1}−{cosA}\:+\:{cos}\:{B}\:−\:{cos}\left({A}+{B}\right)}{\mathrm{1}+{cos}\:{A}\:−\:{cosB}−{cos}\left({A}+{B}\right)}=\:{tan}\frac{{A}}{\mathrm{2}}.{cot}\:\frac{{B}}{\mathrm{2}} \\ $$$$\left.{b}\right)\:{cos}\alpha\:{cos}\left(\mathrm{60}−\alpha\right){cos}\left(\mathrm{60}+\alpha\right)=\:\frac{\mathrm{1}}{\mathrm{4}}{cos}\mathrm{3}\alpha \\ $$$$ \\ $$
Answered by $@ty@m last updated on 11/Jun/18
(a) L.H.S. =((1−cosA + cos B − cos(A+B))/(1+cos A − cosB−cos(A+B))) =((1−2sin ((A+B)/2)sin ((B−A)/2)−1+2sin^2 ((A+B)/2))/(1+2sin ((A+B)/2)sin ((B−A)/2)−1+2sin^2 ((A+B)/2))) =((sin ((A−B)/2)+sin ((A+B)/2))/(sin ((B−A)/2)+sin ((A+B)/2))) =((2sin (A/2)cos (B/2))/(2sin (B/2)cos (A/2))) =tan (A/2)cot (B/2) =R.H.S. (b) L.H.S. = cosα cos(60−α)cos(60+α) =(1/2).cosα.2cos(60−α)cos(60+α) =(1/2).cosα.(cos 120+cos 2α) =(1/2).cosα.(−(1/2)+2cos^2 α−1) =(1/2).cosα.(2cos^2 α−(3/2)) =(1/4).(4cos^3 α−3cos α) =(1/4)cos 3α =R.H.S.
$$\left({a}\right)\:\:{L}.{H}.{S}. \\ $$$$=\frac{\mathrm{1}−{cosA}\:+\:{cos}\:{B}\:−\:{cos}\left({A}+{B}\right)}{\mathrm{1}+{cos}\:{A}\:−\:{cosB}−{cos}\left({A}+{B}\right)} \\ $$$$=\frac{\mathrm{1}−\mathrm{2sin}\:\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{sin}\:\:\frac{{B}−{A}}{\mathrm{2}}−\mathrm{1}+\mathrm{2sin}^{\mathrm{2}} \frac{{A}+{B}}{\mathrm{2}}}{\mathrm{1}+\mathrm{2sin}\:\:\frac{{A}+{B}}{\mathrm{2}}\mathrm{sin}\:\:\frac{{B}−{A}}{\mathrm{2}}−\mathrm{1}+\mathrm{2sin}^{\mathrm{2}} \frac{{A}+{B}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{sin}\:\frac{{A}−{B}}{\mathrm{2}}+\mathrm{sin}\:\frac{{A}+{B}}{\mathrm{2}}}{\mathrm{sin}\:\frac{{B}−{A}}{\mathrm{2}}+\mathrm{sin}\:\frac{{A}+{B}}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2sin}\:\frac{{A}}{\mathrm{2}}\mathrm{cos}\:\frac{{B}}{\mathrm{2}}}{\mathrm{2sin}\:\frac{{B}}{\mathrm{2}}\mathrm{cos}\:\frac{{A}}{\mathrm{2}}} \\ $$$$=\mathrm{tan}\:\frac{{A}}{\mathrm{2}}\mathrm{cot}\:\frac{{B}}{\mathrm{2}} \\ $$$$={R}.{H}.{S}. \\ $$$$\left({b}\right)\:\:{L}.{H}.{S}. \\ $$$$=\:{cos}\alpha\:{cos}\left(\mathrm{60}−\alpha\right){cos}\left(\mathrm{60}+\alpha\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{cos}\alpha.\mathrm{2}{cos}\left(\mathrm{60}−\alpha\right){cos}\left(\mathrm{60}+\alpha\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{cos}\alpha.\left(\mathrm{cos}\:\mathrm{120}+\mathrm{cos}\:\mathrm{2}\alpha\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{cos}\alpha.\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2cos}\:^{\mathrm{2}} \alpha−\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{cos}\alpha.\left(\mathrm{2cos}\:^{\mathrm{2}} \alpha−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}.\left(\mathrm{4cos}\:^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{3}\alpha \\ $$$$={R}.{H}.{S}. \\ $$$$ \\ $$

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