English

If Tan X = − 1 √ 5 and θ Lies in the Iv Quadrant, Then the Value of Cos X is - Mathematics

Advertisements
Advertisements

Question

If tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is

 

Options

  • \[\frac{\sqrt{5}}{\sqrt{6}}\]

     

  • \[\frac{2}{\sqrt{6}}\]

     

  • \[\frac{1}{2}\]

     

  • \[\frac{1}{\sqrt{6}}\]

     

MCQ

Solution

\[\frac{\sqrt{5}}{\sqrt{6}}\]
\[\text{ In the fourth quadrant, }\cos x \text{ and }\sec x\text{ are positive . }\]
\[\cos x = \frac{1}{\sec x}\]
\[ = \frac{1}{\sqrt{\sec^2 x}}\]
\[ = \frac{1}{\sqrt{1 + \tan^2 x}}\]
\[ = \frac{1}{\sqrt{1 + \left( - \frac{1}{\sqrt{5}} \right)^2}}\]
\[ = \frac{1}{\sqrt{\frac{6}{5}}}\]
\[ = \frac{\sqrt{5}}{\sqrt{6}}\]
shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Trigonometric Functions - Exercise 5.5 [Page 41]

APPEARS IN

RD Sharma Mathematics [English] Class 11
Chapter 5 Trigonometric Functions
Exercise 5.5 | Q 9 | Page 41

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

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


Find the general solution of the equation cos 4 x = cos 2 x


If \[\sin x = \frac{a^2 - b^2}{a^2 + b^2}\], then the values of tan x, sec x and cosec x


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 \[m^2 \leq 2\]


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


Prove that: \[\tan\frac{11\pi}{3} - 2\sin\frac{4\pi}{6} - \frac{3}{4} {cosec}^2 \frac{\pi}{4} + 4 \cos^2 \frac{17\pi}{6} = \frac{3 - 4\sqrt{3}}{2}\]

 


Prove that

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

 


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


Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]


If \[\frac{\pi}{2} < x < \frac{3\pi}{2},\text{ then }\sqrt{\frac{1 - \sin x}{1 + \sin x}}\] 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 sec x + tan x = k, cos x =


Find the general solution of the following equation:

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

Find the general solution of the following equation:

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

Solve the following equation:

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

Solve the following equation:

\[\sin x + \sin 2x + \sin 3x + \sin 4x = 0\]

Solve the following equation:

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

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


Solve the following equation:
 cosx + sin x = cos 2x + sin 2x

 


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


Solve the following equation:
\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]


If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.


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


The general solution of the equation \[7 \cos^2 x + 3 \sin^2 x = 4\] is


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


If \[\cot x - \tan x = \sec x\], then, x is equal to

 


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


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


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


Solve the following equations:
2cos 2x – 7 cos x + 3 = 0


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


Choose the correct alternative:
If sin α + cos α = b, then sin 2α is equal to


Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×