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Question-132205

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Question Number 132205 by Arijit last updated on 12/Feb/21
Commented by Arijit last updated on 12/Feb/21
Please Help me to solve this.....
$$\boldsymbol{\mathrm{Please}}\:\boldsymbol{\mathrm{Help}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{this}}….. \\ $$
Answered by mathmax by abdo last updated on 13/Feb/21
A=tan^2 ((π/(16)))+tan^2 (((7π)/(16)))+tan^2 (((2π)/(16)))+tan^2 (((6π)/(16)))+tan^2 (((3π)/(16)))+tan^2 (((5π)/(16))) +tan^2 (((4π)/(16)))=tan^2 ((π/(16)))+tan^2 ((π/2)−(π/(16)))+tan^2 (((2π)/(16)))+tan^2 ((π/2)−((2π)/(16))) +tan^2 (((3π)/(16)))+tan^2 ((π/2)−((3π)/(16)))+tan^2 ((π/4)) =tan^2 ((π/(16)))+(1/(tan^2 ((π/(16))))) +tan^2 (((2π)/(16)))+(1/(tan^2 (((2π)/(16))))) +tan^2 (((3π)/(16)))+(1/(tan^2 (((3π)/(16))))) =tan^2 ((π/(16)))+(1/(tan^2 ((π/(16))))) +((√2)−1)^2 +(1/(((√2)−1)^2 )) +tan^2 (((3π)/(16)))+(1/(tan^2 (((3π)/(16))))) let tan((π/(16)))=x we have tanx =((2tan((x/2)))/(1−tan^2 ((x/2)))) ⇒for x=(π/8) we get (√2)−1 =((2x)/(1−x^2 )) ⇒((√2)−1)(1−x^2 )−2x=0 ⇒ ((√2)−1)+(1−(√2))x^2 −2x=0 ⇒(1−(√2))x^2 −2x+(√2)−1=0 Δ^′ =1−(1−(√2))((√2)−1) =1+((√2)−1)^2 =1+3−2(√2)=4−2(√2) x_1 =((1+(√(4−2(√2))))/(1−(√2)))<0 and x_2 =((1−(√(4−2(√2))))/(1−(√2))) =((−1+(√(4−2(√2))))/( (√2)−1))>0 ⇒ tan((π/(16)))=(((√(4−2(√2)))−1)/( (√2)−1)) =((√2)+1)((√(4−2(√2)))−1) ⇒ tan^2 ((π/(16)))=(3+2(√2))(4−2(√2)+1−2(√(4−2(√2)))) =(3+2(√2))(5−2(√2)−2(√(4−2(√2)))) rest to find tan(((3π)/(16)))...be continued...
$$\mathrm{A}=\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{7}\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{6}\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{5}\pi}{\mathrm{16}}\right) \\ $$$$+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{4}\pi}{\mathrm{16}}\right)=\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{2}\pi}{\mathrm{16}}\right) \\ $$$$+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−\frac{\mathrm{3}\pi}{\mathrm{16}}\right)+\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right) \\ $$$$=\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)}\:+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{16}}\right)+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{2}\pi}{\mathrm{16}}\right)}\:+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{16}}\right)+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{16}}\right)} \\ $$$$=\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)}\:+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }\:+\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{16}}\right)+\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{3}\pi}{\mathrm{16}}\right)} \\ $$$$\mathrm{let}\:\mathrm{tan}\left(\frac{\pi}{\mathrm{16}}\right)=\mathrm{x}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{tanx}\:=\frac{\mathrm{2tan}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right)}\:\Rightarrow\mathrm{for}\:\mathrm{x}=\frac{\pi}{\mathrm{8}}\:\mathrm{we}\:\mathrm{get} \\ $$$$\sqrt{\mathrm{2}}−\mathrm{1}\:=\frac{\mathrm{2x}}{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)−\mathrm{2x}=\mathrm{0}\:\Rightarrow \\ $$$$\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)+\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{2x}=\mathrm{0}\:\Rightarrow\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{2x}+\sqrt{\mathrm{2}}−\mathrm{1}=\mathrm{0} \\ $$$$\Delta^{'} \:=\mathrm{1}−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\:=\mathrm{1}+\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \:=\mathrm{1}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}=\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{1}−\sqrt{\mathrm{2}}}<\mathrm{0}\:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{1}−\sqrt{\mathrm{2}}}\:=\frac{−\mathrm{1}+\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}>\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{tan}\left(\frac{\pi}{\mathrm{16}}\right)=\frac{\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}−\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\:=\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\left(\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}−\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{16}}\right)=\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{1}−\mathrm{2}\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$$$=\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}\sqrt{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}\right)\:\:\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{tan}\left(\frac{\mathrm{3}\pi}{\mathrm{16}}\right)…\mathrm{be}\:\mathrm{continued}… \\ $$
Commented by Arijit last updated on 14/Feb/21
Thank you very much
$${Thank}\:{you}\:{very}\:{much} \\ $$

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