English

Which of the following is incorrect?

Advertisements
Advertisements

Question

Which of the following is incorrect?

Options

  • \[\sin x = - \frac{1}{5}\]

     

  • cos x = 1

  • \[\sec x = \frac{1}{2}\]

     

  • tan x = 20

MCQ
Advertisements

Solution

`bb(sec x = 1/2)`

Explanation:

\[\sin x = - \frac{1}{5}\] is correct as \[- 1 \leq \sin x \leq 1\]

cos x = 1 is correct as cos x = 1 is correct as

\[\sec x = \frac{1}{2}\] is not correct as \[\sec x \in ( - \infty , - 1] \cup [1, \infty )\]
tan x = 20 is correct as tan x can take any real value.
Hence, the correct answer is option \[\sec x = \frac{1}{2}\].
shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Trigonometric Functions - Exercise 5.5 [Page 43]

APPEARS IN

R.D. Sharma Mathematics [English] Class 11
Chapter 5 Trigonometric Functions
Exercise 5.5 | Q 25 | Page 43

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find the principal and general solutions of the equation sec x = 2


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


If \[\cot x \left( 1 + \sin x \right) = 4 m \text{ and }\cot x \left( 1 - \sin x \right) = 4 n,\] \[\left( m^2 + n^2 \right)^2 = mn\]


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

 


In a ∆ABC, prove that:
cos (A + B) + cos C = 0


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


Prove that:

\[\tan\frac{5\pi}{4}\cot\frac{9\pi}{4} + \tan\frac{17\pi}{4}\cot\frac{15\pi}{4} = 0\]

 


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

 

Find the general solution of the following equation:

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

Find the general solution of the following equation:

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

Find the general solution of the following equation:

\[\tan 3x = \cot x\]

Find the general solution of the following equation:

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

Solve the following equation:

\[2 \cos^2 x - 5 \cos x + 2 = 0\]

Solve the following equation:

\[3 \cos^2 x - 2\sqrt{3} \sin x \cos x - 3 \sin^2 x = 0\]

Solve the following equation:

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

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:

`cosec  x = 1 + cot x`


Solve the following equation:
\[\sec x\cos5x + 1 = 0, 0 < x < \frac{\pi}{2}\]


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

 


Solve the following equation:
 sin x tan x – 1 = tan x – sin x

 


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


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


If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.

 

If \[\tan px - \tan qx = 0\], then the values of θ form a series in

 


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


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

 


If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of x is


The number of values of x in the interval [0, 5 π] satisfying the equation \[3 \sin^2 x - 7 \sin x + 2 = 0\] 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 5x − sin x = cos 3


Solve the following equations:
sin θ + sin 3θ + sin 5θ = 0


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


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


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


If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation ______.


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


The minimum value of 3cosx + 4sinx + 8 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×