
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
163.5k+ 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
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

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Instantaneous Velocity - Formula based Examples for JEE

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

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

JEE Advanced 2025 Notes
