y = cot^ 1{a/√(x^2 a^2)} Let x = a sec θ→θ = sec^ 1 ( x/a ) x^2 = a^2 sec^2 θ √(x^2 a^2) = √(a^2 sec^2 θ a^2) = a √(sec^2 θ 1) 1 + tan^2 θ = sec^2 θ ∴ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Substitution Method to Remove Indeterminate Form

-1a/√x2-a2

Question

$c o t^{- 1} {\frac{a}{\sqrt{x^{2} - a^{2}}}}$

Open in App

Solution

$y = c o t^{- 1} {\frac{a}{\sqrt{x^{2} - a^{2}}}}$
$L e t x = a s e c θ \to θ = s e c^{- 1} (\frac{x}{a})$
$x^{2} = a^{2} s e c^{2} θ$
$\sqrt{x^{2} - a^{2}} = \sqrt{a^{2} s e c^{2} θ - a^{2}} = a \sqrt{s e c^{2} θ - 1}$
$1 + t a n^{2} θ = s e c^{2} θ$
$∴ s e c^{2} θ - 1 = t a n^{2} θ$
$∴ \sqrt{x^{2} - a^{2}} = a t a n θ$
$∴ y = c o t^{- 1} {\frac{a}{\sqrt{x^{2} - a^{2}}}} = c o t^{- 1} {\frac{a}{a t a n θ}} = c o t^{- 1} (c o t θ)$
$∴ y = c o t^{- 1} (c o t θ) = θ$
$∴ y = s e c^{- 1} (\frac{x}{a})$

Suggest Corrections

Similar questions

Q. If

e

is eccentricity of ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)

and

e^{'}

is eccentricity of

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a < b)

then

Q. If

\frac{1}{x (x^{2} + a^{2})} = \frac{A}{x} + \frac{B x + C}{x^{2} + a^{2}}

, then

{tan}^{- 1} (\frac{A}{B}) =

Q. Let

y = {sin}^{- 1} \frac{a}{\sqrt{a^{2} + x^{2}}}

, then find

\frac{d y}{d x}

Q. If

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)

and

x^{2} - y^{2} = c^{2}

cut at right angles, then

Q. If

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)

and

x^{2} - y^{2} = c^{2}

cut at right angles, then :

Join BYJU'S Learning Program

cot−1{a√x2−a2}

y=cot−1{a√x2−a2} Let x=a sec θ→θ=sec−1(xa) x2=a2sec2θ √x2−a2=√a2sec2θ−a2=a√sec2θ−1 1+tan2θ=sec2θ ∴ sec2θ−1=tan2θ ∴ √x2−a2=a tan θ ∴ y=cot−1{a√x2−a2}=cot−1{aa tan θ}=cot−1(cot θ) ∴ y=cot−1(cotθ)=θ ∴ y=sec−1(xa)

$c o t^{- 1} {\frac{a}{\sqrt{x^{2} - a^{2}}}}$