
If $\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\cos \left( \dfrac{\pi }{4}\tan \theta \right)$ then $\theta =$.
A. \[n\pi +\dfrac{\pi }{4}\]
B. \[2n\pi \pm \dfrac{\pi }{4}\]
C. \[n\pi -\dfrac{\pi }{4}\]
D. \[2n\pi \pm \dfrac{\pi }{6}\]
Answer
162k+ views
Hint: To find the value of $\theta $ we will first write the given equation in terms of sin. Then comparing both side of the equation we will derive an equation which we will simplify. We will then write the derived equation in terms of tan and again simplify using formula of expansion. We will then apply the theorem which states that if $x,y$ are even multiples of $\dfrac{\pi }{2}$ then $\tan x=\tan y$ implies that $x=n\pi +y$ where $n$is an integer.
Complete step by step solution: We are given a trigonometric equation $\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\cos \left( \dfrac{\pi }{4}\tan \theta \right)$ and we have to determine the value of $\theta $.
We will first write both side of the equation in terms of the trigonometric functions sin.
As we know to convert sin into cos we use the formula $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $. So the equation will be,
$\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\sin \left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{4}\tan \theta \right) \right)$
We will now compare both side of the trigonometric equation. So,
$\dfrac{\pi }{4}\cot \theta =\left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{4}\tan \theta \right) \right)$
We will now simplify the equation.
$\begin{align}
& \dfrac{\pi }{4}\cot \theta =\dfrac{\pi }{2}-\dfrac{\pi }{4}\tan \theta \\
& \dfrac{\pi }{4}\cot \theta +\dfrac{\pi }{4}\tan \theta =\dfrac{\pi }{2} \\
& \dfrac{\pi }{4}\left( \cot \theta +\tan \theta \right)=\dfrac{\pi }{2} \\
& \cot \theta +\tan \theta =2
\end{align}$
We will now write the equation in terms of the trigonometric function tan by using the formula$\cot \theta =\dfrac{1}{\tan \theta }$ and simplify.
\[\begin{align}
& \cot \theta +\tan \theta =2 \\
& \dfrac{1}{\tan \theta }+\tan \theta =2 \\
& \dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }=2 \\
& 1+{{\tan }^{2}}\theta =2\tan \theta \\
& 1+{{\tan }^{2}}\theta -2\tan \theta =0
\end{align}\]
Using expansion formula ${{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}$in the equation above.
\[\begin{align}
& 1+{{\tan }^{2}}\theta -2\tan \theta =0 \\
& {{\left( \tan \theta -1 \right)}^{2}}=0 \\
& \tan \theta -1=0 \\
& \tan \theta =1
\end{align}\]
We know that $\tan \dfrac{\pi }{4}=1$. So,
\[\tan \theta =\tan \dfrac{\pi }{4}\]
Applying the theorem here we will get,
$\theta =n\pi +\dfrac{\pi }{4}$, where $n\in Z$.
The value of $\theta $ for the trigonometric equation $\sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha )$ is $\theta =n\pi +\dfrac{\pi }{4}$
Option ‘A’ is correct
Note: We have used the principal solution of sin after writing the equation in terms of sin that is here $\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\sin \left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{4}\tan \theta \right) \right)$ because we have to further reduce this equation as it has trigonometric function $\cot ,\tan $in it.
Instead of writing the equation $\cot \theta +\tan \theta =2$ in terms of tan we could have also used function sin and cos to simplify the equation and using theorems accordingly.
Complete step by step solution: We are given a trigonometric equation $\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\cos \left( \dfrac{\pi }{4}\tan \theta \right)$ and we have to determine the value of $\theta $.
We will first write both side of the equation in terms of the trigonometric functions sin.
As we know to convert sin into cos we use the formula $\sin \left( \dfrac{\pi }{2}-\theta \right)=\cos \theta $. So the equation will be,
$\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\sin \left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{4}\tan \theta \right) \right)$
We will now compare both side of the trigonometric equation. So,
$\dfrac{\pi }{4}\cot \theta =\left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{4}\tan \theta \right) \right)$
We will now simplify the equation.
$\begin{align}
& \dfrac{\pi }{4}\cot \theta =\dfrac{\pi }{2}-\dfrac{\pi }{4}\tan \theta \\
& \dfrac{\pi }{4}\cot \theta +\dfrac{\pi }{4}\tan \theta =\dfrac{\pi }{2} \\
& \dfrac{\pi }{4}\left( \cot \theta +\tan \theta \right)=\dfrac{\pi }{2} \\
& \cot \theta +\tan \theta =2
\end{align}$
We will now write the equation in terms of the trigonometric function tan by using the formula$\cot \theta =\dfrac{1}{\tan \theta }$ and simplify.
\[\begin{align}
& \cot \theta +\tan \theta =2 \\
& \dfrac{1}{\tan \theta }+\tan \theta =2 \\
& \dfrac{1+{{\tan }^{2}}\theta }{\tan \theta }=2 \\
& 1+{{\tan }^{2}}\theta =2\tan \theta \\
& 1+{{\tan }^{2}}\theta -2\tan \theta =0
\end{align}\]
Using expansion formula ${{a}^{2}}+{{b}^{2}}-2ab={{(a-b)}^{2}}$in the equation above.
\[\begin{align}
& 1+{{\tan }^{2}}\theta -2\tan \theta =0 \\
& {{\left( \tan \theta -1 \right)}^{2}}=0 \\
& \tan \theta -1=0 \\
& \tan \theta =1
\end{align}\]
We know that $\tan \dfrac{\pi }{4}=1$. So,
\[\tan \theta =\tan \dfrac{\pi }{4}\]
Applying the theorem here we will get,
$\theta =n\pi +\dfrac{\pi }{4}$, where $n\in Z$.
The value of $\theta $ for the trigonometric equation $\sin 3\alpha =4\sin \alpha \sin (x+\alpha )\sin (x-\alpha )$ is $\theta =n\pi +\dfrac{\pi }{4}$
Option ‘A’ is correct
Note: We have used the principal solution of sin after writing the equation in terms of sin that is here $\sin \left( \dfrac{\pi }{4}\cot \theta \right)=\sin \left( \dfrac{\pi }{2}-\left( \dfrac{\pi }{4}\tan \theta \right) \right)$ because we have to further reduce this equation as it has trigonometric function $\cot ,\tan $in it.
Instead of writing the equation $\cot \theta +\tan \theta =2$ in terms of tan we could have also used function sin and cos to simplify the equation and using theorems accordingly.
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
