Tan^ 11/√(x^2 1) = tan^ 11/√(cosec^2 θ 1) (Putting x = cosec θ ) tan^ 11/cot θ = θ = cosec^ 1 x. ...

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

Composition of Inverse Trigonometric Functions and Trigonometric Functions

tan-11/√x2 - ...

Question

$t a n^{- 1} \frac{1}{\sqrt{x^{2} - 1}} =$

\frac{x}{2} + c o s e c^{- 1} x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{x}{2} + s e c^{- 1} x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

c o s e c^{- 1} x

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

s e c^{- 1} x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $c o s e c^{- 1} x$
$t a n^{- 1} \frac{1}{\sqrt{x^{2} - 1}} = t a n^{- 1} \frac{1}{\sqrt{c o s e c^{2} θ - 1}}$
(Putting x = cosec $θ$ )
$t a n^{- 1} \frac{1}{c o t θ} = θ = c o s e c^{- 1} x .$

Suggest Corrections

Similar questions

$t a n^{- 1} \frac{1}{\sqrt{x^{2} - 1}} =$

Q. If

y = {tan}^{- 1} \frac{1}{x^{2} + x + 1} + {tan}^{- 1} \frac{1}{x^{2} + 3 x + 3} + {tan}^{- 1} \frac{1}{x^{2} + 5 x + 7} + . . . . .

to n terms, then

Q. If

y = {tan}^{- 1} (\frac{1}{1 + x + x^{2}}) + {tan}^{- 1} (\frac{1}{x^{2} + 3 x + 3}) + {tan}^{- 1} (\frac{1}{x^{2} + 5 x + 7}) + . . . . n

terms then

y^{'} (0)

Q. If

y = {tan}^{- 1} \frac{1}{1 + x + x^{2}} + {tan}^{- 1} \frac{1}{x^{2} + 3 x + 3}

+ {tan}^{- 1} \frac{1}{x^{2} + 5 x + 7} + \dots +

upto

n

terms, then

If $y = {tan}^{- 1} (\frac{1}{1 + x + x^{2}}) + {tan}^{- 1} (\frac{1}{x^{2} + 3 x + 3}) + {tan}^{- 1} (\frac{1}{7 + 5 x + x^{2}}) + \dots n$ terms, then ${(\frac{d y}{d x})}_{x = 0} =$

Join BYJU'S Learning Program

tan−11√x2−1=

The correct option is C cosec−1xtan−11√x2−1=tan−11√cosec2 θ−1 (Putting x = cosec θ ) tan−11cot θ=θ=cosec−1 x.

$t a n^{- 1} \frac{1}{\sqrt{x^{2} - 1}} =$

The correct option is C $c o s e c^{- 1} x$
$t a n^{- 1} \frac{1}{\sqrt{x^{2} - 1}} = t a n^{- 1} \frac{1}{\sqrt{c o s e c^{2} θ - 1}}$
(Putting x = cosec $θ$ )
$t a n^{- 1} \frac{1}{c o t θ} = θ = c o s e c^{- 1} x .$