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Question Number 193533 by horsebrand11 last updated on 16/Jun/23

If cot x−tan x=4 then cot^2 +(2/(sin 2x)) −tan^2 x =?

$$\:\:\:\mathrm{If}\:\mathrm{cot}\:\mathrm{x}−\mathrm{tan}\:\mathrm{x}=\mathrm{4}\:\mathrm{then} \\ $$$$\:\:\:\:\mathrm{cot}\:^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{sin}\:\mathrm{2x}}\:−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\:=? \\ $$

Answered by MM42 last updated on 16/Jun/23

1−tan^2 x=4tanx⇒tan^2 x+4tanx−1=0 tanx=−2±(√5) A=(1/(tan^2 x))+((1+tan^2 x)/(tanx)) −tan^2 x

$$\mathrm{1}−{tan}^{\mathrm{2}} {x}=\mathrm{4}{tanx}\Rightarrow{tan}^{\mathrm{2}} {x}+\mathrm{4}{tanx}−\mathrm{1}=\mathrm{0} \\ $$$${tanx}=−\mathrm{2}\pm\sqrt{\mathrm{5}} \\ $$$${A}=\frac{\mathrm{1}}{{tan}^{\mathrm{2}} {x}}+\frac{\mathrm{1}+{tan}^{\mathrm{2}} {x}}{{tanx}}\:−{tan}^{\mathrm{2}} {x} \\ $$

Answered by Rajpurohith last updated on 19/Jun/23

say a=tanx and b=cotx ⇒ab=1 given b−a=4 ⇒(b+a)^2 −(b−a)^2 =4ab=4 ⇒(b+a)^2 −16=4 ⇒b+a=2(√5) ⇒2b=4+2(√(5 )) ⇒ b=cotx=2+(√5) ⇒a=tanx=−2+(√5) ⇒cot^2 x−tan^2 x=(b+a)(b−a)=2(√5).4=8(√5) so sin2x=((2a)/(1+a^2 ))=((2(√5)−4)/(1+4+5−4(√5)))=((2((√5)−2))/(10−4(√5))) =((2((√5)−2))/( 2(√5)((√5)−2)))=(1/( (√5))) ⇒the value of given expression is (2+(√5))^2 +2(√5)−(−2+(√5))^2 =2(√5)+(2(√5))(4)=10(√5) ■

$${say}\:{a}={tanx}\:{and}\:{b}={cotx}\:\Rightarrow{ab}=\mathrm{1} \\ $$$${given}\:{b}−{a}=\mathrm{4} \\ $$$$\:\Rightarrow\left({b}+{a}\right)^{\mathrm{2}} −\left({b}−{a}\right)^{\mathrm{2}} =\mathrm{4}{ab}=\mathrm{4} \\ $$$$\Rightarrow\left({b}+{a}\right)^{\mathrm{2}} −\mathrm{16}=\mathrm{4}\:\Rightarrow{b}+{a}=\mathrm{2}\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{2}{b}=\mathrm{4}+\mathrm{2}\sqrt{\mathrm{5}\:}\:\Rightarrow\:{b}={cotx}=\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{a}={tanx}=−\mathrm{2}+\sqrt{\mathrm{5}} \\ $$$$\Rightarrow{cot}^{\mathrm{2}} {x}−{tan}^{\mathrm{2}} {x}=\left({b}+{a}\right)\left({b}−{a}\right)=\mathrm{2}\sqrt{\mathrm{5}}.\mathrm{4}=\mathrm{8}\sqrt{\mathrm{5}} \\ $$$${so}\:{sin}\mathrm{2}{x}=\frac{\mathrm{2}{a}}{\mathrm{1}+{a}^{\mathrm{2}} }=\frac{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{4}}{\mathrm{1}+\mathrm{4}+\mathrm{5}−\mathrm{4}\sqrt{\mathrm{5}}}=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)}{\mathrm{10}−\mathrm{4}\sqrt{\mathrm{5}}} \\ $$$$=\frac{\mathrm{2}\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)}{\:\mathrm{2}\sqrt{\mathrm{5}}\left(\sqrt{\mathrm{5}}−\mathrm{2}\right)}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow{the}\:{value}\:{of}\:{given}\:{expression}\:{is} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{5}}−\left(−\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \\ $$$$=\mathrm{2}\sqrt{\mathrm{5}}+\left(\mathrm{2}\sqrt{\mathrm{5}}\right)\left(\mathrm{4}\right)=\mathrm{10}\sqrt{\mathrm{5}}\:\:\:\:\:\:\blacksquare \\ $$$$ \\ $$$$ \\ $$

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