Solve the equation 2tan -1 cos x -tan -1 2 x

2tan^ 1(cosx)=tan^ 1(2cosecx) tan^ 1(2cosx1 cos^2x)=tan^ 1(2cosecx) 2cosxsin^2x=2cosecx cotx=1 x=π4 ...

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Domain and Range of Basic Inverse Trigonometric Functions

Solve the equ...

Question

Solve the equation
$2 {tan}^{- 1} (cos x) = {tan}^{- 1} (2 csc x)$

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2 {tan}^{- 1} (cos x) = {tan}^{- 1} (2 csc x)

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

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

\tan^{- 1}

\sin

x

=

\tan

\sin

x

x \neq \frac{π}{2}

\cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) + \frac{1}{2} \tan^{- 1} (\frac{2 x}{1 - x^{2}}) = \frac{2 π}{3}

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

\tan^{- 1} \frac{1}{4} + 2 \tan^{- 1} \frac{1}{5} + \tan^{- 1} \frac{1}{6} + \tan^{- 1} \frac{1}{x} = \frac{π}{4}

3 \sin^{- 1} \frac{2 x}{1 + x^{2}} - 4 \cos^{- 1} \frac{1 - x^{2}}{1 + x^{2}} + 2 \tan^{- 1} \frac{2 x}{1 - x^{2}} = \frac{π}{3}

\tan^{- 1} (\frac{2 x}{1 - x^{2}}) + \cot^{- 1} (\frac{1 - x^{2}}{2 x}) = \frac{2 π}{3}, x > 0

\tan^{- 1} (\frac{x - 2}{x - 1}) + \tan^{- 1} (\frac{x + 2}{x + 1}) = \frac{π}{4}

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

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