Prove that tan - 1-x - 1-2cos - 1 1 - x1 - x- - 0- 1

RHS = 1 2cos ^ 1( #10;1 x1 + x) #10; #10; 1 2cos ^ 1( 1 #10;tan^2θ1 + tan^2θ) #10; #10; 1 2cos ^ 1( cos #10;2θ) #10 ...

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Properties Derived from Trigonometric Identities

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Question

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

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

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

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{s i n}^{- 1} (1 - x) - 2 {s i n}^{- 1} x = π / 2

{s i n}^{- 1} x + {s i n}^{- 1} (1 - x) = {c o s}^{- 1} x

0, 1 / 2

\tan^{- 1} (\frac{1 - x^{2}}{2 x}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{π}{2}

\sin \{\tan^{- 1} \frac{1 - x^{2}}{2 x} + \cos^{- 1} \frac{1 - x^{2}}{2 x}\} = 1

y = {tan}^{- 1} \frac{u}{\sqrt{1 - u^{2}}}

x = {sec}^{- 1} \frac{1}{2 u^{2} - 1}

u \in (0, \frac{1}{\sqrt{2}})

\cup (\frac{1}{\sqrt{2}}, 1)

2 \frac{d y}{d x} + 1 = 0

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