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Question Number 54755 by Knight last updated on 10/Feb/19
solve for x sin[2cos^(−1) {cot(2tan^(−1) x)}]=0 (x = 1∓(√(2 )) , ∓1 ,−1∓(√2) )
solveforxsin[2cos−1{cot(2tan−1x)}]=0(x=1∓2,∓1,−1∓2)
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Feb/19
θ=tan^(−1) x→tanθ=x cot(2θ)=(1/(tan2θ))=((1−tan^2 θ)/(2tanθ))=((1−x^2 )/(2x)) now sin[2cos^(−1) {cot(2tan^(−1) x)}]=0 sin[2cos^(−1) {cot(2tan^(−1) x)}]=sin0 or sinπ 2cos^(−1) {cot(2tan^(−1) x)}=0 or π cot(2tan^(−1) x)=cos((0/2)) or cos((π/2)) cot(2tan^(−1) x)=1 or 0 ((1−x^2 )/(2x))=1 or ((1−x^2 )/(2x))=0 when 1−x^2 =2x x^2 +2x=1 (x+1)^2 =2 (x+1)=±(√2) x=−1±(√2) when ((1−x^2 )/(2x))=0→1−x^2 =0→x^2 =1 so x=±1
θ=tan−1x→tanθ=xcot(2θ)=1tan2θ=1−tan2θ2tanθ=1−x22xnowsin[2cos−1{cot(2tan−1x)}]=0sin[2cos−1{cot(2tan−1x)}]=sin0orsinπ2cos−1{cot(2tan−1x)}=0orπcot(2tan−1x)=cos(02)orcos(π2)cot(2tan−1x)=1or01−x22x=1or1−x22x=0when1−x2=2xx2+2x=1(x+1)2=2(x+1)=±2x=−1±2when1−x22x=0→1−x2=0→x2=1sox=±1

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