-1-1-x2-1-x-A- -2-1-2 -1 xB- -1 xC- -1-2tan -1 xD- -2-1-2tan -1 x

The correct option is D π2 12tan^ 1x^ 1(√(1+x^2) 1x) Let x=tanθ ^ 1(√(1+tan^2θ) 1tanθ) =^ 1(θ 1tanθ) =^ 1(1 cosθsinθ) =^ 1(tanθ 2) =π2 θ2=π2 12tan^ 1x ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

-1√1+x2-1/x=A...

Question

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

\frac{π}{2} - \frac{1}{2} {cot}^{- 1} x

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{cot}^{- 1} x

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- \frac{1}{2} {tan}^{- 1} x

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\frac{π}{2} - \frac{1}{2} {tan}^{- 1} x

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Solution

The correct option is D $\frac{π}{2} - \frac{1}{2} {tan}^{- 1} x$
${cot}^{- 1} (\frac{\sqrt{1 + x^{2}} - 1}{x})$
Let $x = tan θ$
${cot}^{- 1} (\frac{\sqrt{1 + {tan}^{2} θ} - 1}{tan θ}) = {cot}^{- 1} (\frac{sec θ - 1}{tan θ})$
$= {cot}^{- 1} (\frac{1 - cos θ}{sin θ}) = {cot}^{- 1} (tan \frac{θ}{2}) = \frac{π}{2} - \frac{θ}{2} = \frac{π}{2} - \frac{1}{2} {tan}^{- 1} x$

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Similar questions

Q. If

x \geq 1 t h e n 2 t a n^{- 1} x + s i n^{- 1} (\frac{2 x}{1 + x^{2}})

Q. If x > 1, then

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

is equal to
(a)

4 \tan^{- 1} x

(b) 0
(c)

\frac{π}{2}

(d)

π

Q. Show that

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

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

Q. Assertion :STATEMENT 1: Let

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

f^{'} (2) = - \frac{2}{5}

Reason: STATEMENT 2:

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

\forall x > 1

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cot−1(√1+x2−1x)= _________

The correct option is D π2−12tan−1xcot−1(√1+x2−1x) Let x=tanθ cot−1(√1+tan2θ−1tanθ)=cot−1(secθ−1tanθ) =cot−1(1−cosθsinθ)=cot−1(tanθ2)=π2−θ2=π2−12tan−1x

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