English

If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to ______. - Mathematics

Advertisements
Advertisements

Question

If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to ______.

Options

  • 1

  • 4

  • 2

  • None of these

MCQ
Fill in the Blanks

Solution

If sinθ + cosecθ = 2, then sin2θ + cosec2θ is equal to 2.

Explanation:

sinθ + cosecθ = 2

Squaring L.H.S and R.H.S

We get,

⇒ (sinθ + cosecθ)2 = 22

⇒ (sinθ + cosecθ)2 = 4

⇒ sin2θ + cosec2θ + 2sinθ cosecθ = 4 [∵ `1/sintheta = cosectheta`]

= sin2θ + cosec2θ + 2 = 4

= sin2θ + cosec2θ = 2

shaalaa.com
  Is there an error in this question or solution?
Chapter 3: Trigonometric Functions - Exercise [Page 55]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 11
Chapter 3 Trigonometric Functions
Exercise | Q 30 | Page 55

Video TutorialsVIEW ALL [1]

RELATED QUESTIONS

Find the value of: tan 15°


Prove the following:

sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x


Prove the following:

`(sin x -  siny)/(cos x + cos y)= tan  (x -y)/2`


Prove the following:

`(sin x + sin 3x)/(cos x + cos 3x) = tan 2x`


Prove that: sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x


Prove that: sin 3x + sin 2x – sin x = 4sin x `cos  x/2 cos  (3x)/2`


 If \[\sin A = \frac{12}{13}\text{ and } \sin B = \frac{4}{5}\], where \[\frac{\pi}{2}\] < A < π and 0 < B < \[\frac{\pi}{2}\], find the following:
cos (A + B)


If \[\tan A = \frac{3}{4}, \cos B = \frac{9}{41}\], where π < A < \[\frac{3\pi}{2}\] and 0 < B <\[\frac{\pi}{2}\], find tan (A + B).

 


If \[\cos A = - \frac{12}{13}\text{ and }\cot B = \frac{24}{7}\], where A lies in the second quadrant and B in the third quadrant, find the values of the following:
cos (A + B)


Prove that
\[\frac{\tan A + \tan B}{\tan A - \tan B} = \frac{\sin \left( A + B \right)}{\sin \left( A - B \right)}\]


Prove that

\[\frac{\cos 8^\circ - \sin 8^\circ}{\cos 8^\circ + \sin 8^\circ} = \tan 37^\circ\]

Prove that:

\[\sin\left( \frac{\pi}{3} - x \right)\cos\left( \frac{\pi}{6} + x \right) + \cos\left( \frac{\pi}{3} - x \right)\sin\left( \frac{\pi}{6} + x \right) = 1\]

 


Prove that:

\[\sin\left( \frac{4\pi}{9} + 7 \right)\cos\left( \frac{\pi}{9} + 7 \right) - \cos\left( \frac{4\pi}{9} + 7 \right)\sin\left( \frac{\pi}{9} + 7 \right) = \frac{\sqrt{3}}{2}\]

 


Prove that \[\frac{\tan 69^\circ + \tan 66^\circ}{1 - \tan 69^\circ \tan 66^\circ} = - 1\].


Prove that:
sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.


Prove that:
\[\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0\]

 


Prove that:

\[\frac{\sin \left( A - B \right)}{\sin A \sin B} + \frac{\sin \left( B - C \right)}{\sin B \sin C} + \frac{\sin \left( C - A \right)}{\sin C \sin A} = 0\]

 


Prove that:
tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x


If tan A = x tan B, prove that
\[\frac{\sin \left( A - B \right)}{\sin \left( A + B \right)} = \frac{x - 1}{x + 1}\]


If \[\tan\theta = \frac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}\] , then show that \[\sin\alpha + \cos\alpha = \sqrt{2}\cos\theta\].


Write the maximum value of 12 sin x − 9 sin2 x


Write the interval in which the value of 5 cos x + 3 cos \[\left( x + \frac{\pi}{3} \right) + 3\] lies. 


If \[\frac{\cos \left( x - y \right)}{\cos \left( x + y \right)} = \frac{m}{n}\]  then write the value of tan x tan y


The value of \[\sin^2 \frac{5\pi}{12} - \sin^2 \frac{\pi}{12}\] 


If \[\tan A = \frac{a}{a + 1}\text{ and } \tan B = \frac{1}{2a + 1}\] 


Express the following as the sum or difference of sines and cosines:
2 sin 4x sin 3x


If sinθ + cosθ = 1, then find the general value of θ.


Find the most general value of θ satisfying the equation tan θ = –1 and cos θ = `1/sqrt(2)`.


3(sinx – cosx)4 + 6(sinx + cosx)2 + 4(sin6x + cos6x) = ______.


State whether the statement is True or False? Also give justification.

If cosecx = 1 + cotx then x = 2nπ, 2nπ + `pi/2`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×