1
Question
Prove that:
tan−1[√1+x−√1−x√1+x+√1−x]=π4−12cos−1x,−1√2≤x≤1
tan−1[√1+x−√1−x√1+x+√1−x]=π4−12cos−1x,−1√2≤x≤1
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Solution
Put x=cosθ in the LHS of tan−1[√1+x−√1−x√1+x+√1−x]=π4−12cos−1x.
tan−1[√1+cosθ−√1−cosθ√1+cosθ+√1−cosθ]
=tan−1⎡⎢
⎢
⎢
⎢⎣√2cos2θ2−√2sin2θ2√2cos2θ2+√2sin2θ2⎤⎥
⎥
⎥
⎥⎦
=tan−1⎡⎢
⎢
⎢⎣√2cosθ2−√2sinθ2√2cosθ2+√2sinθ2⎤⎥
⎥
⎥⎦
=tan−1⎡⎢
⎢
⎢
⎢⎣√2(cosθ2−sinθ2)√2(cosθ2+sinθ2)⎤⎥
⎥
⎥
⎥⎦
=tan−1⎡⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢
⎢⎣⎛⎜
⎜
⎜⎝cosθ2cosθ2−sinθ2cosθ2⎞⎟
⎟
⎟⎠⎛⎜
⎜
⎜⎝cosθ2cosθ2+sinθ2cosθ2⎞⎟
⎟
⎟⎠⎤⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
=tan−1⎡⎢
⎢
⎢
⎢⎣(1−tanθ2)(1+tanθ2)⎤⎥
⎥
⎥
⎥⎦
=tan−1⎡⎢
⎢
⎢
⎢⎣tanπ4−tanθ21+(tanπ4)(tanθ2)⎤⎥
⎥
⎥
⎥⎦
=tan−1[tan(π4−θ2)]
=π4−θ2
As x=cosθ, then θ=cos−1x.
=π4−12cos−1x
=RHS
Put x=cosθ in the LHS of tan−1[√1+x−√1−x√1+x+√1−x]=π4−12cos−1x.
tan−1[√1+cosθ−√1−cosθ√1+cosθ+√1−cosθ]
=tan−1⎡⎢ ⎢ ⎢ ⎢⎣√2cos2θ2−√2sin2θ2√2cos2θ2+√2sin2θ2⎤⎥ ⎥ ⎥ ⎥⎦
=tan−1⎡⎢ ⎢ ⎢⎣√2cosθ2−√2sinθ2√2cosθ2+√2sinθ2⎤⎥ ⎥ ⎥⎦
=tan−1⎡⎢ ⎢ ⎢ ⎢⎣√2(cosθ2−sinθ2)√2(cosθ2+sinθ2)⎤⎥ ⎥ ⎥ ⎥⎦
=tan−1⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣⎛⎜ ⎜ ⎜⎝cosθ2cosθ2−sinθ2cosθ2⎞⎟ ⎟ ⎟⎠⎛⎜ ⎜ ⎜⎝cosθ2cosθ2+sinθ2cosθ2⎞⎟ ⎟ ⎟⎠⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦
=tan−1⎡⎢ ⎢ ⎢ ⎢⎣(1−tanθ2)(1+tanθ2)⎤⎥ ⎥ ⎥ ⎥⎦
=tan−1⎡⎢ ⎢ ⎢ ⎢⎣tanπ4−tanθ21+(tanπ4)(tanθ2)⎤⎥ ⎥ ⎥ ⎥⎦
=tan−1[tan(π4−θ2)]
=π4−θ2
As x=cosθ, then θ=cos−1x.
=π4−12cos−1x
=RHS
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