Tan ( sec^ 1√(1+x^2) ) = tan ( sec^ 1√(1+tan^2 θ) ) (Putting x = tan θ) = tan(sec^ 1 sec θ) = tan θ = x. ...

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

Composition of Trigonometric Functions and Inverse Trigonometric Functions

tan [ sec-1√1...

Question

$t a n [s e c^{- 1} \sqrt{1 + x^{2}}] =$

$\frac{1}{x}$

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

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

\frac{1}{\sqrt{1 + x^{2}}}

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

$\frac{x}{\sqrt{1 + x^{2}}}$

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

Open in App

Solution

The correct option is B
x

$t a n (s e c^{- 1} \sqrt{1 + x^{2}}) = t a n (s e c^{- 1} \sqrt{1 + t a n^{2} θ})$
(Putting $x = t a n θ$ )
$= t a n (s e c^{- 1} s e c θ) = t a n θ = x$ .

Suggest Corrections

Similar questions

$t a n [s e c^{- 1} \sqrt{1 + x^{2}}] =$

\frac{sec θ + 1 - tan θ}{sec θ + 1 + tan θ} + \frac{tan θ + sec θ - 1}{tan θ - sec θ + 1} = 2 sec θ

\begin{matrix} F u n c t i o n & D e r i v a t i v e s a) s i n^{- 1} x & 1) \frac{- 1}{| x | \sqrt{x^{2} - 1}} b) c o s^{- 1} x & 2) \frac{- 1}{1 + x^{2}} c) t a n^{- 1} x & 3) \frac{1}{| x | \sqrt{x^{2} - 1}} d) s e c^{- 1} x & 4) \frac{1}{1 + x^{2}} e) c o t^{- 1} x & 5) \frac{- 1}{\sqrt{1 - x^{2}}} f) c o s e c^{- 1} x & 6) \frac{1}{\sqrt{1 - x^{2}}} \end{matrix}

\frac{tan θ}{sec θ - 1} = \frac{tan θ + sec θ + 1}{tan θ + sec θ - 1}

\frac{\tan A}{s e c A - 1} = \frac{\tan A + s e c A + 1}{\tan A + s e c A - 1}

Join BYJU'S Learning Program