12tan-1 x - cos11 - -1 - x22-1 - x212-

1 2 tan ^ 1 x =cos ^ 1 #10;{ 1+√( 1+ x ^ 2 ) #160; 2√( 1+ x ^ 2 #10;) #160; #160; } #160; ^ 1 #160; 2 #160; #10; #10;R. ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Domain and Range of Basic Inverse Trigonometric Functions

12tan-1 x = c...

Question

$\frac{1}{2} {tan}^{- 1} x = {cos}^{1} {\frac{1 + \sqrt{1 + x^{2}}}{2 \sqrt{1 + x^{2}}}}^{\frac{1}{2}}$ .

Open in App

Solution

Suggest Corrections

Similar questions

\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}

$\int \frac{x^{2} + 1}{x^{4} + 1} d x$ will be equal to which of the following

\int \frac{(x^{2} + 1) d x}{(x^{2} + 2) (2 x^{2} + 1)} = \frac{1}{3 \sqrt{(2)}} {{tan}^{- 1} \frac{x}{\sqrt{(2)}} + {tan}^{- 1} x \sqrt{2}} .

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

2 {tan}^{- 1} x .

\int 2 {tan}^{- 1} x = 2 [x {tan}^{- 1} x - \frac{1}{2} log (1 + x^{2})]

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

Join BYJU'S Learning Program