English

Find the General Solution of the Following Equation: Tan M X + Cot N X = 0 - Mathematics

Advertisements
Advertisements

Question

Find the general solution of the following equation:

\[\tan mx + \cot nx = 0\]
Sum
Advertisements

Solution

We have:

\[\tan mx + \cot nx = 0\]

\[\Rightarrow \tan mx = - \cot nx\]

\[ \Rightarrow \tan mx = \tan \left( \frac{\pi}{2} + nx \right)\]

\[ \Rightarrow mx = r\pi + \left( \frac{\pi}{2} + nx \right), r \in Z\]

\[ \Rightarrow (m - n) x = r\pi + \frac{\pi}{2}, r \in Z\]

\[ \Rightarrow x = \left( \frac{2r + 1}{m - n} \right)\frac{\pi}{2}, r \in Z\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Trigonometric equations - Exercise 11.1 [Page 21]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 11 Trigonometric equations
Exercise 11.1 | Q 2.08 | Page 21

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find the principal and general solutions of the equation  `cot x = -sqrt3`


If \[\sin x + \cos x = m\], then prove that \[\sin^6 x + \cos^6 x = \frac{4 - 3 \left( m^2 - 1 \right)^2}{4}\], where \[m^2 \leq 2\]


Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]


If \[T_n = \sin^n x + \cos^n x\], prove that \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]


Prove that: tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0


Prove that cos 570° sin 510° + sin (−330°) cos (−390°) = 0

Prove that:
\[\sec\left( \frac{3\pi}{2} - x \right)\sec\left( x - \frac{5\pi}{2} \right) + \tan\left( \frac{5\pi}{2} + x \right)\tan\left( x - \frac{3\pi}{2} \right) = - 1 .\]


In a ∆ABC, prove that:

\[\tan\frac{A + B}{2} = \cot\frac{C}{2}\]

In a ∆A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that cos(180° − A) + cos (180° + B) + cos (180° + C) − sin (90° + D) = 0


Prove that:
\[\sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} = \frac{1}{2}\]


If \[\frac{3\pi}{4} < \alpha < \pi, \text{ then }\sqrt{2\cot \alpha + \frac{1}{\sin^2 \alpha}}\] is equal to


The value of \[\tan1^\circ \tan2^\circ \tan3^\circ . . . \tan89^\circ\] is

 

Find the general solution of the following equation:

\[cosec x = - \sqrt{2}\]

Find the general solution of the following equation:

\[\sqrt{3} \sec x = 2\]

Find the general solution of the following equation:

\[\sin 2x = \frac{\sqrt{3}}{2}\]

Find the general solution of the following equation:

\[\sin 9x = \sin x\]

Find the general solution of the following equation:

\[\sin 2x + \cos x = 0\]

Solve the following equation:
\[\sin^2 x - \cos x = \frac{1}{4}\]


Solve the following equation:

\[\cos x + \cos 2x + \cos 3x = 0\]

Solve the following equation:

\[\tan x + \tan 2x = \tan 3x\]

Solve the following equation:

\[\tan 3x + \tan x = 2\tan 2x\]

Solve the following equation:

\[\sqrt{3} \cos x + \sin x = 1\]


Solve the following equation:
3 – 2 cos x – 4 sin x – cos 2x + sin 2x = 0


In (0, π), the number of solutions of the equation ​ \[\tan x + \tan 2x + \tan 3x = \tan x \tan 2x \tan 3x\] is 


General solution of \[\tan 5 x = \cot 2 x\] is


Find the principal solution and general solution of the following:
cot θ = `sqrt(3)`


Find the principal solution and general solution of the following:
tan θ = `- 1/sqrt(3)`


Solve the following equations for which solution lies in the interval 0° ≤ θ < 360°

sin4x = sin2x


Solve the following equations:
sin θ + cos θ = `sqrt(2)`


Solve the following equations:
cot θ + cosec θ = `sqrt(3)`


Solve the following equations:
cos 2θ = `(sqrt(5) + 1)/4`


Choose the correct alternative:
`(cos 6x + 6 cos 4x + 15cos x + 10)/(cos 5x + 5cs 3x + 10 cos x)` is equal to


Solve `sqrt(3)` cos θ + sin θ = `sqrt(2)`


If 2sin2θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×