Question-196725

Question Number 196725 by sonukgindia last updated on 30/Aug/23

Answered by Frix last updated on 30/Aug/23

(c) t^8 +(1/t^8 )=m t^8 +(1/t^8 )+2=m+2 (t^4 +(1/t^4 ))^2 =m+2 t^4 +(1/t^4 )=(√(m+2)) Same procedure again.. t^4 +(1/t^4 )+2=(√(m+2))+2 t^2 +(1/t^2 )=(√((√(m+2))+2)) ...and again t+(1/t)=(√((√((√(m+2))+2))+2))

$$\left({c}\right) \\ $$$${t}^{\mathrm{8}} +\frac{\mathrm{1}}{{t}^{\mathrm{8}} }={m} \\ $$$${t}^{\mathrm{8}} +\frac{\mathrm{1}}{{t}^{\mathrm{8}} }+\mathrm{2}={m}+\mathrm{2} \\ $$$$\left({t}^{\mathrm{4}} +\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\right)^{\mathrm{2}} ={m}+\mathrm{2} \\ $$$${t}^{\mathrm{4}} +\frac{\mathrm{1}}{{t}^{\mathrm{4}} }=\sqrt{{m}+\mathrm{2}} \\ $$$$\mathrm{Same}\:\mathrm{procedure}\:\mathrm{again}.. \\ $$$${t}^{\mathrm{4}} +\frac{\mathrm{1}}{{t}^{\mathrm{4}} }+\mathrm{2}=\sqrt{{m}+\mathrm{2}}+\mathrm{2} \\ $$$${t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }=\sqrt{\sqrt{{m}+\mathrm{2}}+\mathrm{2}} \\ $$$$…\mathrm{and}\:\mathrm{again} \\ $$$${t}+\frac{\mathrm{1}}{{t}}=\sqrt{\sqrt{\sqrt{{m}+\mathrm{2}}+\mathrm{2}}+\mathrm{2}} \\ $$

Answered by AST last updated on 30/Aug/23

[{(tanθ+cotθ)^2 −2}^2 −2]^2 −2=m ⇒tanθ+cotθ=(√((√((√(m+2))+2))+2))

$$\left[\left\{\left({tan}\theta+{cot}\theta\right)^{\mathrm{2}} −\mathrm{2}\right\}^{\mathrm{2}} −\mathrm{2}\right]^{\mathrm{2}} −\mathrm{2}={m} \\ $$$$\Rightarrow{tan}\theta+{cot}\theta=\sqrt{\sqrt{\sqrt{{m}+\mathrm{2}}+\mathrm{2}}+\mathrm{2}} \\ $$

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Question-196725

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