हिंदी

If X = R Sin θ Cos ϕ, Y = R Sin θ Sin ϕ and Z = R Cos θ, Then X2 + Y2 + Z2 is Independent of

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • θ, ϕ

  • r, θ

  • r, ϕ

  • r

MCQ
Advertisements

उत्तर

θ, ϕ
We have:
x = r sin θ cos ϕ  ,  y = r sin θ sin ϕ and z = r cos θ,
∴ x2 + y2 + z2

\[= \left( r \sin\theta \cos\phi \right)^2 + \left( r \sin\theta \sin\phi \right)^2 + \left( r \cos\theta \right)^2 \]

\[ = r^2 \sin^2 \theta \cos^2 \phi + r^2 \sin^2 \theta \sin^2 \phi + r^2 \cos^2 \theta \]

\[ = r^2 \sin^2 \theta \left( \cos^2 \phi + \sin^2 \phi \right) + r^2 \cos^2 \theta \]

\[ = r^2 \sin^2 \theta \times 1 + r^2 \cos^2 \theta\]

\[ = r^2 \sin^2 \theta + r^2 \cos^2 \theta\]

\[ = r^2 \left( \sin^2 \theta + \cos^2 \theta \right)\]

\[ = r^2 \times 1\]

\[ = r^2 \]

\[\text{ Thus, }x^2 + y^2 + z^2\text{ is independent of }\theta\text{ and }\phi .\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Trigonometric Functions - Exercise 5.5 [पृष्ठ ४१]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 5 Trigonometric Functions
Exercise 5.5 | Q 7 | पृष्ठ ४१

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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


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


If \[x = \frac{2 \sin x}{1 + \cos x + \sin x}\], then prove that

\[\frac{1 - \cos x + \sin x}{1 + \sin x}\] is also equal to a.

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


Prove that

\[\frac{\sin(180^\circ + x) \cos(90^\circ + x) \tan(270^\circ - x) \cot(360^\circ - x)}{\sin(360^\circ - x) \cos(360^\circ + x) cosec( - x) \sin(270^\circ + x)} = 1\]

 


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:
cos (A + B) + cos C = 0


In a ∆ABC, prove that:

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

If sec \[x = x + \frac{1}{4x}\], then sec x + tan x = 

 

If \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] 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 tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is

 

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


Find the general solution of the following equation:

\[\sin 2x = \cos 3x\]

Find the general solution of the following equation:

\[\tan x + \cot 2x = 0\]

Solve the following equation:

\[\tan^2 x + \left( 1 - \sqrt{3} \right) \tan x - \sqrt{3} = 0\]

Solve the following equation:

\[\cos 4 x = \cos 2 x\]

Solve the following equation:

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

Solve the following equation:

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

Solve the following equation:

\[\sin 3x - \sin x = 4 \cos^2 x - 2\]

Solve the following equation:

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

Solve the following equation:

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


Solve the following equation:

\[\sin x + \cos x = 1\]

Solve the following equation:
\[\cot x + \tan x = 2\]

 


Solve the following equation:
\[5 \cos^2 x + 7 \sin^2 x - 6 = 0\]


Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]


Solve the following equation:
3tanx + cot x = 5 cosec x


Solve the following equation:
3sin2x – 5 sin x cos x + 8 cos2 x = 2


Write the number of points of intersection of the curves

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

Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]

 and cos 2x are in A.P.


Write the solution set of the equation 

\[\left( 2 \cos x + 1 \right) \left( 4 \cos x + 5 \right) = 0\] in the interval [0, 2π].

A solution of the equation \[\cos^2 x + \sin x + 1 = 0\], lies in the interval


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

 

The solution of the equation \[\cos^2 x + \sin x + 1 = 0\] lies in the interval


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

sin4x = sin2x


Solve the equation sin θ + sin 3θ + sin 5θ = 0


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×