
If $y = \sin^{ - 1}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + \sec^{ - 1}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$. Then what is the value of $\dfrac{{dy}}{{dx}}$?
A. $\dfrac{4}{{\left( {1 - {x^2}} \right)}}$
B. $\dfrac{4}{{\left( {1 + {x^2}} \right)}}$
C. $\dfrac{1}{{\left( {1 + {x^2}} \right)}}$
D. $\dfrac{{ - 4}}{{\left( {1 + {x^2}} \right)}}$
Answer
164.1k+ views
Hint: Solve the given trigonometric equation by substituting $x = \tan\theta $. Further simplify the equation using the identities of $\sin 2A$ and $\cos 2A$. Then substitute the value of $\theta $ and simplify the equation. In the end, differentiate the equation with respect to $x$ and get the required answer.
Formula Used:
$\sin 2A = \dfrac{{2\tan A}}{{1 + \tan^{2}A}}$
$\cos 2A = \dfrac{{1 + \tan^{2}A}}{{1 - \tan^{2}A}}$
Complete step by step solution:
The given trigonometric equation is $y = \sin^{ - 1}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + \sec^{ - 1}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$.
Let’s simplify the above equation.
Substitute $x = \tan\theta $ in the above equation.
$y = \sin^{ - 1}\left( {\dfrac{{2\tan\theta }}{{1 + \tan^{2}\theta }}} \right) + \sec^{ - 1}\left( {\dfrac{{1 + \tan^{2}\theta }}{{1 - \tan^{2}\theta }}} \right)$
$ \Rightarrow y = \sin^{ - 1}\left( {\dfrac{{2\tan\theta }}{{1 + \tan^{2}\theta }}} \right) + \cos^{ - 1}\left( {\dfrac{{1 - \tan^{2}\theta }}{{1 + \tan^{2}\theta }}} \right)$ $\because \sec^{ - 1}A = \cos^{ - 1}\left( {\dfrac{1}{A}} \right)$
Now apply the formulas of $\sin 2A$, and $\cos 2A$.
We get,
$y = \sin^{ - 1}\left( {\sin2\theta } \right) + \cos^{ - 1}\left( {\cos2\theta } \right)$
Now apply the inverse trigonometric rules $\sin^{ - 1}\left( {\sin A} \right) = A$, and $\cos^{ - 1}\left( {\cos A} \right) = A$.
$y = 2\theta + 2\theta $
$ \Rightarrow y = 4\theta $
Resubstitute the value of $\theta $.
$y = 4\tan^{ - 1}x$
Differentiate the above equation with respect to $x$.
$\dfrac{{dy}}{{dx}} = 4\dfrac{d}{{dx}}\left( {\tan^{ - 1}x} \right)$
Apply the formula of a derivative $\dfrac{d}{{dx}}\left( {\tan^{ - 1}x} \right) = \dfrac{1}{{1 + {x^2}}}$.
$\dfrac{{dy}}{{dx}} = 4\left( {\dfrac{1}{{1 + {x^2}}}} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{\left( {1 + {x^2}} \right)}}$
Option ‘B’ is correct
Note: The first principle of differentiation can also be used to help solve the problem. Trigonometric identities and formulas are used to simplify the function in the questions, and basic differentiation results are then used to determine the right answer. When changing a variable's value, we must be careful.
Formula Used:
$\sin 2A = \dfrac{{2\tan A}}{{1 + \tan^{2}A}}$
$\cos 2A = \dfrac{{1 + \tan^{2}A}}{{1 - \tan^{2}A}}$
Complete step by step solution:
The given trigonometric equation is $y = \sin^{ - 1}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right) + \sec^{ - 1}\left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$.
Let’s simplify the above equation.
Substitute $x = \tan\theta $ in the above equation.
$y = \sin^{ - 1}\left( {\dfrac{{2\tan\theta }}{{1 + \tan^{2}\theta }}} \right) + \sec^{ - 1}\left( {\dfrac{{1 + \tan^{2}\theta }}{{1 - \tan^{2}\theta }}} \right)$
$ \Rightarrow y = \sin^{ - 1}\left( {\dfrac{{2\tan\theta }}{{1 + \tan^{2}\theta }}} \right) + \cos^{ - 1}\left( {\dfrac{{1 - \tan^{2}\theta }}{{1 + \tan^{2}\theta }}} \right)$ $\because \sec^{ - 1}A = \cos^{ - 1}\left( {\dfrac{1}{A}} \right)$
Now apply the formulas of $\sin 2A$, and $\cos 2A$.
We get,
$y = \sin^{ - 1}\left( {\sin2\theta } \right) + \cos^{ - 1}\left( {\cos2\theta } \right)$
Now apply the inverse trigonometric rules $\sin^{ - 1}\left( {\sin A} \right) = A$, and $\cos^{ - 1}\left( {\cos A} \right) = A$.
$y = 2\theta + 2\theta $
$ \Rightarrow y = 4\theta $
Resubstitute the value of $\theta $.
$y = 4\tan^{ - 1}x$
Differentiate the above equation with respect to $x$.
$\dfrac{{dy}}{{dx}} = 4\dfrac{d}{{dx}}\left( {\tan^{ - 1}x} \right)$
Apply the formula of a derivative $\dfrac{d}{{dx}}\left( {\tan^{ - 1}x} \right) = \dfrac{1}{{1 + {x^2}}}$.
$\dfrac{{dy}}{{dx}} = 4\left( {\dfrac{1}{{1 + {x^2}}}} \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{4}{{\left( {1 + {x^2}} \right)}}$
Option ‘B’ is correct
Note: The first principle of differentiation can also be used to help solve the problem. Trigonometric identities and formulas are used to simplify the function in the questions, and basic differentiation results are then used to determine the right answer. When changing a variable's value, we must be careful.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets
