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Question Number 117852 by bemath last updated on 14/Oct/20

Determine the value of   (1)(tan ((7π)/(24))+tan ((5π)/(24))).cos (π/(12)) + 2 .  (2) (((9−4(√5))/(5x)))^(1/(4 ))  .(5(√x) +(√(20x)) )^(0.5) . 2^(−1)  = ?

$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{tan}\:\frac{\mathrm{7}\pi}{\mathrm{24}}+\mathrm{tan}\:\frac{\mathrm{5}\pi}{\mathrm{24}}\right).\mathrm{cos}\:\frac{\pi}{\mathrm{12}}\:+\:\mathrm{2}\:. \\ $$$$\left(\mathrm{2}\right)\:\sqrt[{\mathrm{4}\:}]{\frac{\mathrm{9}−\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{5x}}}\:.\left(\mathrm{5}\sqrt{\mathrm{x}}\:+\sqrt{\mathrm{20x}}\:\right)^{\mathrm{0}.\mathrm{5}} .\:\mathrm{2}^{−\mathrm{1}} \:=\:? \\ $$

Answered by bobhans last updated on 14/Oct/20

(2) (((9−2(√(20)))/(5x)))^(1/(4 ))  = (((((√5)−2)^2 )/(5x)))^(1/(4 )) = ((√((√5)−2))/( ((5x))^(1/(4 )) ))  ⇒(((9−4(√5))/(5x)))^(1/(4 ))  .(5(√x)+(√(20x)))^(0.5)  .2^(−1)  =  ⇒((√((√5)−2))/( ((5x))^(1/(4 )) )) . (√(5(√x)+2(√(5x)))) .(1/2) =   ⇒((√(((√5)−2)(5(√x)+2(√(5x)))))/(2 ((5x))^(1/(4 )) )) = ((√(5(√(5x))+10(√x)−10(√x)−4(√(5x))))/(2 ((5x))^(1/(4 )) ))  = ((√(√(5x)))/(2 ((5x))^(1/(4 )) )) = (1/2)

$$\left(\mathrm{2}\right)\:\sqrt[{\mathrm{4}\:}]{\frac{\mathrm{9}−\mathrm{2}\sqrt{\mathrm{20}}}{\mathrm{5x}}}\:=\:\sqrt[{\mathrm{4}\:}]{\frac{\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{5x}}}=\:\frac{\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\:\sqrt[{\mathrm{4}\:}]{\mathrm{5x}}} \\ $$$$\Rightarrow\sqrt[{\mathrm{4}\:}]{\frac{\mathrm{9}−\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{5x}}}\:.\left(\mathrm{5}\sqrt{\mathrm{x}}+\sqrt{\mathrm{20x}}\right)^{\mathrm{0}.\mathrm{5}} \:.\mathrm{2}^{−\mathrm{1}} \:= \\ $$$$\Rightarrow\frac{\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\:\sqrt[{\mathrm{4}\:}]{\mathrm{5x}}}\:.\:\sqrt{\mathrm{5}\sqrt{\mathrm{x}}+\mathrm{2}\sqrt{\mathrm{5x}}}\:.\frac{\mathrm{1}}{\mathrm{2}}\:=\: \\ $$$$\Rightarrow\frac{\sqrt{\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)\left(\mathrm{5}\sqrt{\mathrm{x}}+\mathrm{2}\sqrt{\mathrm{5x}}\right)}}{\mathrm{2}\:\sqrt[{\mathrm{4}\:}]{\mathrm{5x}}}\:=\:\frac{\sqrt{\mathrm{5}\sqrt{\mathrm{5x}}+\mathrm{10}\sqrt{\mathrm{x}}−\mathrm{10}\sqrt{\mathrm{x}}−\mathrm{4}\sqrt{\mathrm{5x}}}}{\mathrm{2}\:\sqrt[{\mathrm{4}\:}]{\mathrm{5x}}} \\ $$$$=\:\frac{\sqrt{\sqrt{\mathrm{5x}}}}{\mathrm{2}\:\sqrt[{\mathrm{4}\:}]{\mathrm{5x}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by bobhans last updated on 14/Oct/20

(1) letting x = (π/(24))  ⇒(tan 7x+tan 5x).cos 2x +2 =  ⇒(((sin 7x.cos 5x+cos 7xsin 5x)/(cos 7x cos 5x))).cos 2x +2 =  ⇒((sin 12x)/(2cos 7x cos 5x)). 2cos 2x+2 =  ⇒((2cos 2x)/(cos 12x+cos 2x)) + 2 ; [ sin 12x = sin (π/2)=1 ]  [ & cos 12x = cos (π/2)=0 ]  ⇒ ((2cos (π/(12)))/(cos (π/(12)))) + 2 = 4

$$\left(\mathrm{1}\right)\:\mathrm{letting}\:\mathrm{x}\:=\:\frac{\pi}{\mathrm{24}} \\ $$$$\Rightarrow\left(\mathrm{tan}\:\mathrm{7x}+\mathrm{tan}\:\mathrm{5x}\right).\mathrm{cos}\:\mathrm{2x}\:+\mathrm{2}\:= \\ $$$$\Rightarrow\left(\frac{\mathrm{sin}\:\mathrm{7x}.\mathrm{cos}\:\mathrm{5x}+\mathrm{cos}\:\mathrm{7xsin}\:\mathrm{5x}}{\mathrm{cos}\:\mathrm{7x}\:\mathrm{cos}\:\mathrm{5x}}\right).\mathrm{cos}\:\mathrm{2x}\:+\mathrm{2}\:= \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\mathrm{12x}}{\mathrm{2cos}\:\mathrm{7x}\:\mathrm{cos}\:\mathrm{5x}}.\:\mathrm{2cos}\:\mathrm{2x}+\mathrm{2}\:= \\ $$$$\Rightarrow\frac{\mathrm{2cos}\:\mathrm{2x}}{\mathrm{cos}\:\mathrm{12x}+\mathrm{cos}\:\mathrm{2x}}\:+\:\mathrm{2}\:;\:\left[\:\mathrm{sin}\:\mathrm{12x}\:=\:\mathrm{sin}\:\frac{\pi}{\mathrm{2}}=\mathrm{1}\:\right] \\ $$$$\left[\:\&\:\mathrm{cos}\:\mathrm{12x}\:=\:\mathrm{cos}\:\frac{\pi}{\mathrm{2}}=\mathrm{0}\:\right] \\ $$$$\Rightarrow\:\frac{\mathrm{2cos}\:\frac{\pi}{\mathrm{12}}}{\mathrm{cos}\:\frac{\pi}{\mathrm{12}}}\:+\:\mathrm{2}\:=\:\mathrm{4} \\ $$

Answered by mindispower last updated on 14/Oct/20

tg(a)+tg(b)=((sin(a+b))/(cos(a)cos(b)))=((2sin(a+b))/(cos(a−b)+cos(a+b)))  ⇒tg(((7π)/(24)))+tg(((5π)/(24)))=((2sin((π/2)))/(cos((π/(12)))+cos((π/2))))  (2/(cos((π/(12)))))  1⇔((2/(cos((π/(12))))))cos((π/(12)))+2=4

$${tg}\left({a}\right)+{tg}\left({b}\right)=\frac{{sin}\left({a}+{b}\right)}{{cos}\left({a}\right){cos}\left({b}\right)}=\frac{\mathrm{2}{sin}\left({a}+{b}\right)}{{cos}\left({a}−{b}\right)+{cos}\left({a}+{b}\right)} \\ $$$$\Rightarrow{tg}\left(\frac{\mathrm{7}\pi}{\mathrm{24}}\right)+{tg}\left(\frac{\mathrm{5}\pi}{\mathrm{24}}\right)=\frac{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{2}}\right)}{{cos}\left(\frac{\pi}{\mathrm{12}}\right)+{cos}\left(\frac{\pi}{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{2}}{{cos}\left(\frac{\pi}{\mathrm{12}}\right)} \\ $$$$\mathrm{1}\Leftrightarrow\left(\frac{\mathrm{2}}{{cos}\left(\frac{\pi}{\mathrm{12}}\right)}\right){cos}\left(\frac{\pi}{\mathrm{12}}\right)+\mathrm{2}=\mathrm{4} \\ $$

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