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Prove that :- { ^{ - 1}} x = {1}{2}{ ^{ - 1}}( {{{1 - x}}{{1 +

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Prove that :-
$tan^{- 1} x = \frac{1}{2} cos^{- 1} (\frac{1 - x}{1 + x}), x \in [0, 1]$

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Prove that :-

tan^{- 1} x = \frac{1}{2} cos^{- 1} (\frac{1 - x}{1 + x}), x \in [0, 1]

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Text solutionVerified

Taking R.H.S

\frac{1}{2} co s^{- 1} (\frac{1 - x}{1 + x})

putting

x . t a n^{2} θ

= \frac{1}{2} co s^{- 1} (\frac{1 - t a n ^{2} θ}{1 + t a n ^{2} θ})

= \frac{1}{2} (\frac{1 - \frac{s i n ^{2} θ}{co s ^{2} θ}}{1 + \frac{s i n ^{2} θ}{co s ^{2} θ}})

= \frac{1}{2} co s^{- 1} (\frac{\frac{co s ^{2} θ - s i n ^{2} θ}{co s ^{2} θ}}{\frac{co s ^{2} θ + s i n ^{2} θ}{co s ^{2} θ}})

= \frac{1}{2} co s^{- 1} (\frac{co s ^{2} θ - s i n ^{2} θ}{co s ^{2} θ} \times \frac{co s ^{2} θ}{s i n ^{2} θ + co s ^{2} θ})

= \frac{1}{2} . co s^{- 1} (\frac{co s ^{2} θ . s i n ^{2} θ}{s i n ^{2} θ + co s ^{2} θ})

= \frac{1}{2} . co s^{- 1} (\frac{co s ^{2} θ - s i n ^{2} θ}{1})

[

∴

using

s i n^{2} θ + co s^{2} θ = 1]

\frac{1}{2} . co s^{- 1} (cos 2 θ)

(using

cos 2 θ - co s^{2} θ - s i n^{2} θ)

= \frac{1}{2} . θ

= θ

we assume that

x = t a n^{2} θ

x = t an θ

t a n^{- 1} x = θ

Hence,

\frac{1}{2} . co s^{- 1} (\frac{1 - x}{1 + x}) = 0

\frac{1}{2} . co s^{- 1} (\frac{1 - x}{1 + x}) = t a n^{- 1} x

L.H.S = R.H.S

Hence proved.

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Question Text	Prove that :- $tan^{- 1} x = \frac{1}{2} cos^{- 1} (\frac{1 - x}{1 + x}), x \in [0, 1]$
Answer Type	Text solution:1
Upvotes	150