मराठी

Find the General Solution of the Following Equation: Tan P X = Cot Q X

Advertisements
Advertisements

प्रश्न

Find the general solution of the following equation:

\[\tan px = \cot qx\]

 

बेरीज
Advertisements

उत्तर

We have:

\[\tan px = \cot qx\]

\[\Rightarrow \tan px = \tan \left( \frac{\pi}{2} - qx \right)\]

\[ \Rightarrow px = n\pi + \left( \frac{\pi}{2} - qx \right), n \in Z\]

\[ \Rightarrow (p + q)x = n\pi + \frac{\pi}{2}, n \in Z\]

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Trigonometric equations - Exercise 11.1 [पृष्ठ २१]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
पाठ 11 Trigonometric equations
Exercise 11.1 | Q 2.09 | पृष्ठ २१

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the general solution of cosec x = –2


Find the general solution of the equation cos 3x + cos x – cos 2x = 0


Find the general solution for each of the following equations sec2 2x = 1– tan 2x


If \[\tan x = \frac{b}{a}\] , then find the values of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\].


If \[\tan x = \frac{a}{b},\] show that

\[\frac{a \sin x - b \cos x}{a \sin x + b \cos x} = \frac{a^2 - b^2}{a^2 + b^2}\]

If \[a = \sec x - \tan x \text{ and }b = cosec x + \cot x\], then shown that  \[ab + a - b + 1 = 0\]


Prove that:  tan 225° cot 405° + tan 765° cot 675° = 0


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


Find x from the following equations:
\[x \cot\left( \frac{\pi}{2} + \theta \right) + \tan\left( \frac{\pi}{2} + \theta \right)\sin \theta + cosec\left( \frac{\pi}{2} + \theta \right) = 0\]


Prove that:

\[\sin\frac{10\pi}{3}\cos\frac{13\pi}{6} + \cos\frac{8\pi}{3}\sin\frac{5\pi}{6} = - 1\]

\[\sqrt{\frac{1 + \cos x}{1 - \cos x}}\] is equal to

 


If \[\frac{\pi}{2} < x < \pi, \text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}}\] is equal to


If x = r sin θ cos ϕ, y = r sin θ sin ϕ and r cos θ, then x2 + y2 + z2 is independent of


If tan A + cot A = 4, then tan4 A + cot4 A is equal to


If x sin 45° cos2 60° = \[\frac{\tan^2 60^\circ cosec30^\circ}{\sec45^\circ \cot^{2^\circ} 30^\circ}\], then x =

 

If \[cosec x + \cot x = \frac{11}{2}\], then tan x =

 


If \[f\left( x \right) = \cos^2 x + \sec^2 x\], then


Solve the following equation:

\[\sin x + \sin 5x = \sin 3x\]

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:
\[\cot x + \tan x = 2\]

 


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


Write the number of points of intersection of the curves

\[2y = 1\] and \[y = \cos x, 0 \leq x \leq 2\pi\].
 

Write the number of values of x in [0, 2π] that satisfy the equation \[\sin x - \cos x = \frac{1}{4}\].


If \[3\tan\left( x - 15^\circ \right) = \tan\left( x + 15^\circ \right)\] \[0 < x < 90^\circ\], find θ.


The smallest value of x satisfying the equation

\[\sqrt{3} \left( \cot x + \tan x \right) = 4\] is 

If a is any real number, the number of roots of \[\cot x - \tan x = a\] in the first quadrant is (are).


The smallest positive angle which satisfies the equation ​

\[2 \sin^2 x + \sqrt{3} \cos x + 1 = 0\] is

A value of x satisfying \[\cos x + \sqrt{3} \sin x = 2\] is

 

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


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

2 sin2x + 1 = 3 sin x


Solve the following equations:
sin 2θ – cos 2θ – sin θ + cos θ = θ


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


Solve the following equations:
`tan theta + tan (theta + pi/3) + tan (theta + (2pi)/3) = sqrt(3)`


Choose the correct alternative:
If cos pθ + cos qθ = 0 and if p ≠ q, then θ is equal to (n is any integer)


Solve 2 tan2x + sec2x = 2 for 0 ≤ x ≤ 2π.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×