1
Question
Solve the equation tan−1[1−x1+x]=12tan−1x,(x>0)
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Solution
We have,tan−1[1−x1+x]=12tan−1x2tan−11−x1+x=tan−1x→(i)
We know that,2tan−1x=tan−1(2x1−x2)2tan−1(1−x1+x)=tan−1⎡⎢
⎢
⎢
⎢
⎢⎣2(1−x1+x)1−(1−x1+x)2⎤⎥
⎥
⎥
⎥
⎥⎦=tan−1⎡⎢
⎢
⎢
⎢
⎢
⎢⎣2(1−x1+x)(1+x)2−(1−x)2(1+x)2⎤⎥
⎥
⎥
⎥
⎥
⎥⎦=tan−1[2(1−x)(1+x)(1+x)2−(1−x)2]using(a+b)(a−b)=a2−b2=tan−1[2(1−x2)(1+x+1−x)(1+x−1+x)]=tan−1[2(1−x2)4x]=tan−1[1−x22x]2tan−1(1−x1+x)=tan−1[1−x22x]fromequation(i)2tan−1(1−x1+x)=tan−1xtan−1(1−x22x)=tan−1x1−x22x=11−x2=2x21−x2−2x2=01−3x2=03x2=03x2=1x=±1√3
x=−1√3 is not possible because x>0
Taken only x=1√3
Hence, this is the answer.
tan−1[1−x1+x]=12tan−1x2tan−11−x1+x=tan−1x→(i)
We know that,
2tan−1x=tan−1(2x1−x2)2tan−1(1−x1+x)=tan−1⎡⎢
⎢
⎢
⎢
⎢⎣2(1−x1+x)1−(1−x1+x)2⎤⎥
⎥
⎥
⎥
⎥⎦=tan−1⎡⎢
⎢
⎢
⎢
⎢
⎢⎣2(1−x1+x)(1+x)2−(1−x)2(1+x)2⎤⎥
⎥
⎥
⎥
⎥
⎥⎦=tan−1[2(1−x)(1+x)(1+x)2−(1−x)2]using(a+b)(a−b)=a2−b2=tan−1[2(1−x2)(1+x+1−x)(1+x−1+x)]=tan−1[2(1−x2)4x]=tan−1[1−x22x]2tan−1(1−x1+x)=tan−1[1−x22x]fromequation(i)2tan−1(1−x1+x)=tan−1xtan−1(1−x22x)=tan−1x1−x22x=11−x2=2x21−x2−2x2=01−3x2=03x2=03x2=1x=±1√3
x=−1√3 is not possible because x>0
Taken only x=1√3
Hence, this is the answer.
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