
Find the value of \[\dfrac{d}{{dx}}\sqrt {\dfrac{{1 - \sin 2x}}{{1 + \sin 2x}}} \] .
A. \[{\sec ^2}x\]
B. \[ - {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\]
C. \[{\sec ^2}\left( {\dfrac{\pi }{4} + x} \right)\]
D. \[{\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\]
Answer
162k+ views
Hint:
While watching the trigonometric equation inside the square root it looks very complex hence our first step will be to simplify the terms using the basic trigonometric formula and then we will be differentiating the term using chain rule.
Formula Used:
\[\sin 2x = 2\sin x\cos x\],
\[{\cos ^2}x + {\sin ^2}x = 1\] ,
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\],
\[\dfrac{d}{{dx}}[f(g(x))] = f\prime (g(x))g\prime (x)\]
Complete step-by-step answer:
By using the Double Angle Property we will solve the trigonometric equation inside square root.
Then \[\sin 2x\] can be written as \[2\sin x\cos x\] .
\[\dfrac{d}{{dx}}\sqrt {\dfrac{{1 - \sin 2x}}{{1 + \sin 2x}}} \]
\[ \Rightarrow \sqrt {\dfrac{{1 - 2\sin x\cos x}}{{1 + 2\sin x\cos x}}} \]
By using the formula \[{\cos ^2}x + {\sin ^2}x = 1\] we will substitute the value of 1 inside the square root.
\[ \Rightarrow \sqrt {\dfrac{{{{\cos }^2}x + {{\sin }^2}x - 2\sin x\cos x}}{{{{\cos }^2}x + {{\sin }^2}x + 2\sin x\cos x}}} \]
By applying the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] we can simplify the terms.
\[ \Rightarrow \sqrt {\dfrac{{{{\left( {\cos x - \sin x} \right)}^2}}}{{{{\left( {\cos x + \sin x} \right)}^2}}}} \]
\[ \Rightarrow \sqrt {{{\left( {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\left( {\cos x + \sin x} \right)}}} \right)}^2}} \]
\[ \Rightarrow \left( {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\left( {\cos x + \sin x} \right)}}} \right)\]
Dividing Numerator and Denominator both by \[\cos x\] to make the terms more simpler,
\[ \Rightarrow \left( {\dfrac{{\left( {\dfrac{{\cos x}}{{\cos x}}} \right) - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{{\cos x}}{{\cos x}}} \right) + \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}} \right)\]
\[ \Rightarrow \left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)\]
According to the property we can write \[1\] as \[\tan \dfrac{\pi }{4}\]
\[ \Rightarrow \left( {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan x}}{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan x}}} \right)\]
According to the formula \[ \Rightarrow \left( {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan x}}{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan x}}} \right)\] can be written as \[\tan \left( {\dfrac{\pi }{4} - x} \right)\]
\[ \Rightarrow \tan \left( {\dfrac{\pi }{4} - x} \right)\]
We have got the inner trigonometric values in the simplest form and now we will be differentiating it,
\[\dfrac{d}{{dx}}\left( {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right)\]
By Applying chain rule we will be differentiating the term.
\[\dfrac{d}{{dx}}[f(g(x))] = f\prime (g(x))g\prime (x)\]
Where \[f\left( x \right) = \tan x\] and \[g\left( x \right) = \left( {\dfrac{\pi }{4} - x} \right)\]
Applying the formula and differentiating the term we will get ,
\[\dfrac{d}{{dx}}\left( {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right)\]
Differentiation of \[\tan x = {\sec ^2}x\]
\[ \Rightarrow {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right).\dfrac{d}{{dx}}\left( {\dfrac{\pi }{4} - x} \right)\]
Differentiation of \[\dfrac{d}{{dx}}\left( {\dfrac{\pi }{4} - x} \right) = \left( { - 1} \right)\]
\[ \Rightarrow - {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\]
Hence the answer is (B) which is\[ - {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\].
Note:
Students generally make mistakes while using the trigonometric formula as well while using the differentiation via chain rule hence students must be very careful while doing the differentiation .
While watching the trigonometric equation inside the square root it looks very complex hence our first step will be to simplify the terms using the basic trigonometric formula and then we will be differentiating the term using chain rule.
Formula Used:
\[\sin 2x = 2\sin x\cos x\],
\[{\cos ^2}x + {\sin ^2}x = 1\] ,
\[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\],
\[\dfrac{d}{{dx}}[f(g(x))] = f\prime (g(x))g\prime (x)\]
Complete step-by-step answer:
By using the Double Angle Property we will solve the trigonometric equation inside square root.
Then \[\sin 2x\] can be written as \[2\sin x\cos x\] .
\[\dfrac{d}{{dx}}\sqrt {\dfrac{{1 - \sin 2x}}{{1 + \sin 2x}}} \]
\[ \Rightarrow \sqrt {\dfrac{{1 - 2\sin x\cos x}}{{1 + 2\sin x\cos x}}} \]
By using the formula \[{\cos ^2}x + {\sin ^2}x = 1\] we will substitute the value of 1 inside the square root.
\[ \Rightarrow \sqrt {\dfrac{{{{\cos }^2}x + {{\sin }^2}x - 2\sin x\cos x}}{{{{\cos }^2}x + {{\sin }^2}x + 2\sin x\cos x}}} \]
By applying the formula \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\] we can simplify the terms.
\[ \Rightarrow \sqrt {\dfrac{{{{\left( {\cos x - \sin x} \right)}^2}}}{{{{\left( {\cos x + \sin x} \right)}^2}}}} \]
\[ \Rightarrow \sqrt {{{\left( {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\left( {\cos x + \sin x} \right)}}} \right)}^2}} \]
\[ \Rightarrow \left( {\dfrac{{\left( {\cos x - \sin x} \right)}}{{\left( {\cos x + \sin x} \right)}}} \right)\]
Dividing Numerator and Denominator both by \[\cos x\] to make the terms more simpler,
\[ \Rightarrow \left( {\dfrac{{\left( {\dfrac{{\cos x}}{{\cos x}}} \right) - \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}{{\left( {\dfrac{{\cos x}}{{\cos x}}} \right) + \left( {\dfrac{{\sin x}}{{\cos x}}} \right)}}} \right)\]
\[ \Rightarrow \left( {\dfrac{{1 - \tan x}}{{1 + \tan x}}} \right)\]
According to the property we can write \[1\] as \[\tan \dfrac{\pi }{4}\]
\[ \Rightarrow \left( {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan x}}{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan x}}} \right)\]
According to the formula \[ \Rightarrow \left( {\dfrac{{\tan \left( {\dfrac{\pi }{4}} \right) - \tan x}}{{\tan \left( {\dfrac{\pi }{4}} \right) + \tan x}}} \right)\] can be written as \[\tan \left( {\dfrac{\pi }{4} - x} \right)\]
\[ \Rightarrow \tan \left( {\dfrac{\pi }{4} - x} \right)\]
We have got the inner trigonometric values in the simplest form and now we will be differentiating it,
\[\dfrac{d}{{dx}}\left( {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right)\]
By Applying chain rule we will be differentiating the term.
\[\dfrac{d}{{dx}}[f(g(x))] = f\prime (g(x))g\prime (x)\]
Where \[f\left( x \right) = \tan x\] and \[g\left( x \right) = \left( {\dfrac{\pi }{4} - x} \right)\]
Applying the formula and differentiating the term we will get ,
\[\dfrac{d}{{dx}}\left( {\tan \left( {\dfrac{\pi }{4} - x} \right)} \right)\]
Differentiation of \[\tan x = {\sec ^2}x\]
\[ \Rightarrow {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right).\dfrac{d}{{dx}}\left( {\dfrac{\pi }{4} - x} \right)\]
Differentiation of \[\dfrac{d}{{dx}}\left( {\dfrac{\pi }{4} - x} \right) = \left( { - 1} \right)\]
\[ \Rightarrow - {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\]
Hence the answer is (B) which is\[ - {\sec ^2}\left( {\dfrac{\pi }{4} - x} \right)\].
Note:
Students generally make mistakes while using the trigonometric formula as well while using the differentiation via chain rule hence students must be very careful while doing the differentiation .
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

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

JoSAA JEE Main & Advanced 2025 Counselling: Registration Dates, Documents, Fees, Seat Allotment & Cut‑offs

NIT Cutoff Percentile for 2025

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

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

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
