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Question Number 83680 by Power last updated on 05/Mar/20

Commented by john santu last updated on 05/Mar/20

considering A+B = 30^o B = 30^o −A ⇒ tan B = (((1/(√3))−tan A)/(1+(1/(√3)) tan A)) tan B = ((1−(√3) tan A)/((√3)+tan A)) (√3) + tan B = (√3)+((1−(√3) tan A)/((√3)+tan A)) = (4/((√3)+tan A)) now ((√3)+tan A)((√3)+tan B) = ((√3)+tan A)((4/((√3)+tan A))) = 4 so ⇒ (4^(14) )((√3) +tan 15^o ) i think it easy

$$\mathrm{considering}\:\mathrm{A}+\mathrm{B}\:=\:\mathrm{30}^{\mathrm{o}} \\ $$$$\mathrm{B}\:=\:\mathrm{30}^{\mathrm{o}} −\mathrm{A}\:\Rightarrow\:\mathrm{tan}\:\mathrm{B}\:=\:\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}−\mathrm{tan}\:\mathrm{A}}{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\mathrm{tan}\:\mathrm{A}} \\ $$$$\mathrm{tan}\:\mathrm{B}\:=\:\frac{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{A}}{\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}} \\ $$$$\sqrt{\mathrm{3}}\:+\:\mathrm{tan}\:\mathrm{B}\:=\:\sqrt{\mathrm{3}}+\frac{\mathrm{1}−\sqrt{\mathrm{3}}\:\mathrm{tan}\:\mathrm{A}}{\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}} \\ $$$$=\:\frac{\mathrm{4}}{\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}} \\ $$$$\mathrm{now}\:\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}\right)\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{B}\right)\:= \\ $$$$\left(\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}\right)\left(\frac{\mathrm{4}}{\sqrt{\mathrm{3}}+\mathrm{tan}\:\mathrm{A}}\right)\:=\:\mathrm{4}\: \\ $$$$\mathrm{so}\:\Rightarrow\:\left(\mathrm{4}^{\mathrm{14}} \:\right)\left(\sqrt{\mathrm{3}}\:+\mathrm{tan}\:\mathrm{15}^{\mathrm{o}} \right)\: \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{it}\:\mathrm{easy} \\ $$$$ \\ $$

Commented by jagoll last updated on 05/Mar/20

tan 15^o = ((1−(1/(√3)))/(1+(1/(√3)))) = (((√3)−1)/((√3)+1))= ((4−2(√3))/2) = 2−(√3) finally = 4^(14) ((√3)+2−(√3)) = 4^(14) ×2 sorry mistake

$$\mathrm{tan}\:\mathrm{15}^{\mathrm{o}} \:=\:\frac{\mathrm{1}−\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}{\mathrm{1}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}}\:=\:\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\sqrt{\mathrm{3}}+\mathrm{1}}=\:\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$=\:\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\mathrm{finally}\:=\:\mathrm{4}^{\mathrm{14}} \left(\sqrt{\mathrm{3}}+\mathrm{2}−\sqrt{\mathrm{3}}\right)\:=\:\mathrm{4}^{\mathrm{14}} ×\mathrm{2} \\ $$$$\mathrm{sorry}\:\mathrm{mistake} \\ $$

Commented by MJS last updated on 05/Mar/20

not 14×4 but 4^(14) ⇒ answer is 2×4^(14) =2^(29) =536 870 912

$$\mathrm{not}\:\mathrm{14}×\mathrm{4}\:\mathrm{but}\:\mathrm{4}^{\mathrm{14}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{2}×\mathrm{4}^{\mathrm{14}} =\mathrm{2}^{\mathrm{29}} =\mathrm{536}\:\mathrm{870}\:\mathrm{912} \\ $$

Commented by john santu last updated on 05/Mar/20

oo yes sir. 4×4×4×...×4 (14 times)

$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{sir}.\:\mathrm{4}×\mathrm{4}×\mathrm{4}×...×\mathrm{4}\:\left(\mathrm{14}\:\mathrm{times}\right) \\ $$

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