
General solution of \[\tan 5\theta = cot2\theta \]
A. \[\theta = \dfrac{{n\pi }}{7} + \dfrac{\pi }{{14}}\]
B. \[\theta = \dfrac{{n\pi }}{7} + \dfrac{\pi }{5}\]
C. \[\theta = \dfrac{{n\pi }}{7} + \dfrac{\pi }{2}\]
D. \[\theta = \dfrac{{n\pi }}{7} + \dfrac{\pi }{3}\]
Answer
162.9k+ views
Hint: Using the knowledge that the answer to the equation \[tan{\rm{ }}x = tan\] is given by \[x = n + \]for any arbitrary integer n, we apply the complementary angle relation \[tan = cot\left( 2 \right)\]to convert the cotangent in the preceding equation \[tan{\rm{ }}5 = cot{\rm{ }}2\]to tangent and then discover the solution.
Complete step by step solution:

We are aware that in a right-angled triangle, the horizontal side is known as the base and is denoted by the letter \[b\]. The vertical side is known as the perpendicular and is denoted by the letter \[p\].
Here, in the right-angled triangle ABC figure above, we have:
\[AC = h,AB = p,BC = b\]
We can infer from trigonometric ratios that the opposite side to the adjacent side (excluding the hypotenuse, also known as leg adjacent) is the tangent of a right-angled triangle. As a result, we have the angle's tangent \[\theta \].
\[tan\theta = ABAC = pb\]
The term "co-tangent" refers to the ratio of the leg next to the opposing side, which is indicated by the symbol \[\cot \theta \] and is provided by
\[cot\theta = AC/AB = bp\]
We are aware that a triangle's total number of angles is \[180 \circ \]. Thus, we have
\[A + B + C = {180^ \circ }\]
\[ \Rightarrow A + C = {90^ \circ }\]
\[ \Rightarrow A = {90^ \circ } - C = {90^ \circ } - \theta \]
For angles A and C, we apply the definitions of tangent and cotangent,
\[tanC = tan\theta = pb,cotC = cot\theta = bp\]
\[cotA = cot(90 - \theta ) = pb,cotA = cot(90 - \theta ) = bp\]
\[tan(90 \circ - \theta ) = cot\theta ,cot(90 \circ - \theta ) = sin\theta \]
The above mentioned formula is also known as the tangent-cotangent reduction formula or complementary angle relation of tangent and cotangent.
We are aware of the solutions to the equation \[\tan x = \tan \alpha \], where x is the unknown and \[\alpha \] is the known angle.
\[x = n\pi + \alpha \]
The trigonometric equation with an unknown angle is provided in the question as,
\[tan5\theta = cot2\theta \]
\[ \Rightarrow tan5\theta = tan(90 \circ - 2\theta )\]
\[ \Rightarrow tan5\theta = tan(\pi /2 - 2\theta )\]
\[\alpha = \dfrac{\pi }{2} - 2\theta \]
\[ \Rightarrow 5\theta = n\pi + \pi /2 - 2\theta \]
\[ \Rightarrow 7\theta = n\pi + \pi /2\]
To get the desired outcome, we divide the aforementioned calculation by seven.
\[ \Rightarrow \theta = n\pi /7 + \pi /14,n \in Z\]
Option ‘A’ is correct
Note: The given result is well defined because the tangent function is not defined for \[\theta = \left( {2n + 1} \right)\dfrac{\pi }{2}\]and the cotangent function is not defined for\[\theta = n\pi \]. The complimentary and supplementary angle relations, which are determined by a shift of radians, must not be confused.\[\tan \left( {\pi + \theta } \right) = - \tan \theta ,\cot \left( {\pi + \theta } \right) = - \cot \theta \].
Complete step by step solution:

We are aware that in a right-angled triangle, the horizontal side is known as the base and is denoted by the letter \[b\]. The vertical side is known as the perpendicular and is denoted by the letter \[p\].
Here, in the right-angled triangle ABC figure above, we have:
\[AC = h,AB = p,BC = b\]
We can infer from trigonometric ratios that the opposite side to the adjacent side (excluding the hypotenuse, also known as leg adjacent) is the tangent of a right-angled triangle. As a result, we have the angle's tangent \[\theta \].
\[tan\theta = ABAC = pb\]
The term "co-tangent" refers to the ratio of the leg next to the opposing side, which is indicated by the symbol \[\cot \theta \] and is provided by
\[cot\theta = AC/AB = bp\]
We are aware that a triangle's total number of angles is \[180 \circ \]. Thus, we have
\[A + B + C = {180^ \circ }\]
\[ \Rightarrow A + C = {90^ \circ }\]
\[ \Rightarrow A = {90^ \circ } - C = {90^ \circ } - \theta \]
For angles A and C, we apply the definitions of tangent and cotangent,
\[tanC = tan\theta = pb,cotC = cot\theta = bp\]
\[cotA = cot(90 - \theta ) = pb,cotA = cot(90 - \theta ) = bp\]
\[tan(90 \circ - \theta ) = cot\theta ,cot(90 \circ - \theta ) = sin\theta \]
The above mentioned formula is also known as the tangent-cotangent reduction formula or complementary angle relation of tangent and cotangent.
We are aware of the solutions to the equation \[\tan x = \tan \alpha \], where x is the unknown and \[\alpha \] is the known angle.
\[x = n\pi + \alpha \]
The trigonometric equation with an unknown angle is provided in the question as,
\[tan5\theta = cot2\theta \]
\[ \Rightarrow tan5\theta = tan(90 \circ - 2\theta )\]
\[ \Rightarrow tan5\theta = tan(\pi /2 - 2\theta )\]
\[\alpha = \dfrac{\pi }{2} - 2\theta \]
\[ \Rightarrow 5\theta = n\pi + \pi /2 - 2\theta \]
\[ \Rightarrow 7\theta = n\pi + \pi /2\]
To get the desired outcome, we divide the aforementioned calculation by seven.
\[ \Rightarrow \theta = n\pi /7 + \pi /14,n \in Z\]
Option ‘A’ is correct
Note: The given result is well defined because the tangent function is not defined for \[\theta = \left( {2n + 1} \right)\dfrac{\pi }{2}\]and the cotangent function is not defined for\[\theta = n\pi \]. The complimentary and supplementary angle relations, which are determined by a shift of radians, must not be confused.\[\tan \left( {\pi + \theta } \right) = - \tan \theta ,\cot \left( {\pi + \theta } \right) = - \cot \theta \].
Recently Updated Pages
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

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals 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
