Answer
385.8k+ views
Hint:We know that$\csc x = \dfrac{1}{{\sin x}}$. So first we need to find $\sin \left( {\dfrac{{11\pi }}{6}}
\right)$and then find its reciprocal.
We also know$\;\cos 2\theta = 1 - 2{\sin ^2}\theta $, this is one of the basic trigonometric identities.
In order to solve this question we can use the above mentioned identity. For that we have to convert our question in such a way that it can be expressed in the form of the above given identity.
Complete step by step solution:
Given
$\csc \left( {\dfrac{{11\pi }}{6}} \right)...................................\left( i \right)$
Now we also know$\csc x = \dfrac{1}{{\sin x}}$. So to find the value of $\csc \left( {\dfrac{{11\pi }}{6}}
\right)$we need to find $\sin \left( {\dfrac{{11\pi }}{6}} \right)$and then find it’s reciprocal.
Now to find $\sin \left( {\dfrac{{11\pi }}{6}} \right)$we can use the identity$\;\cos 2\theta = 1 - 2{\sin
^2}\theta $.
Finding the value of$\sin \left( {\dfrac{{11\pi }}{6}} \right)$:
Now let’s assume $\sin \left( {\dfrac{{11\pi }}{6}} \right) = \sin a......................\left( {ii} \right)$
So similarly we can write $\;\cos a = \cos \left( {\dfrac{{11\pi }}{6}} \right)$
$ \Rightarrow \cos 2a = \cos \left( {\dfrac{{22\pi }}{6}} \right)$
We have to find the value of $\cos \left( {\dfrac{{22\pi }}{6}} \right)$such that by using the identity we can then solve the question.
So finding the value of$\cos \left( {\dfrac{{22\pi }}{6}} \right)$:
We know that $\cos \left( {\dfrac{{22\pi }}{6}} \right)$can be written as
$
\cos \left( {\dfrac{{12\left( {2\pi } \right)}}{6} - \dfrac{{2\pi }}{6}} \right) = \cos \left( {2\left( {2\pi }
\right) - \dfrac{{2\pi }}{6}} \right).................(iii) \\
\\
$
So from (iii) we know that \[\cos \left( {2\left( {2\pi } \right) - \dfrac{{2\pi }}{6}} \right)\]would be in
the IV Quadrant where cosine is positive.
Such that:
\[\cos \left( {2\left( {2\pi } \right) - \dfrac{{2\pi }}{6}} \right) = \cos \left( {\dfrac{{2\pi }}{6}}
\right)..................(iv)\]
Also we know \[\cos \left( {\dfrac{{2\pi }}{6}} \right) = \cos \left( {\dfrac{\pi }{3}} \right) =
\dfrac{1}{2}....................(v)\]
Now by using the identity $\;\cos 2\theta = 1 - 2{\sin ^2}\theta $we get
$
\Rightarrow \cos 2a = 1 - 2{\sin ^2}a = \dfrac{1}{2} \\
\Rightarrow 2{\sin ^2}a = 1 - \dfrac{1}{2} = \dfrac{1}{2} \\
\Rightarrow {\sin ^2}a = \dfrac{1}{4} \\
$
Now from (ii) \[\sin \left( {\dfrac{{11\pi }}{6}} \right) = \sin a\]
$
\Rightarrow \sin {\left( {\dfrac{{11\pi }}{6}} \right)^2} = \dfrac{1}{4} \\
\Rightarrow \sin \left( {\dfrac{{11\pi }}{6}} \right) = - \dfrac{1}{2} \\
$
Now we know that$\csc x = \dfrac{1}{{\sin x}}$, such that:
$\csc \left( {\dfrac{{11\pi }}{6}} \right) = \dfrac{1}{{\sin \left( {\dfrac{{11\pi }}{6}} \right)}} =
\dfrac{1}{{\left( { - \dfrac{1}{2}} \right)}} = - 2$
Therefore the value of$\csc \left( {\dfrac{{11\pi }}{6}} \right)\;{\text{is}}\; - 2$.
Note:
General things to be known for solving this question.
I Quadrant:$0\; - \;\dfrac{\pi }{2}$ All values are positive.
II Quadrant:$\dfrac{\pi }{2}\; - \;\pi $ Only Sine and Cosec values are positive.
III Quadrant:$\pi \; - \;\dfrac{{3\pi }}{2}$ Only Tan and Cot values are positive.
IV Quadrant:$\dfrac{{3\pi }}{2}\; - \;2\pi $ Only Cos and Sec values are positive.
Some other equations needed for solving these types of problem are:
\[
\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right) \\
\cos \left( {2\theta } \right) = {\cos ^2}\left( \theta \right)-{\sin ^2}\left( \theta \right) = 1-2{\text{
}}{\sin ^2}\left( \theta \right) = 2{\text{ }}{\cos ^2}\left( \theta \right)-1 \\
\]
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly.
\right)$and then find its reciprocal.
We also know$\;\cos 2\theta = 1 - 2{\sin ^2}\theta $, this is one of the basic trigonometric identities.
In order to solve this question we can use the above mentioned identity. For that we have to convert our question in such a way that it can be expressed in the form of the above given identity.
Complete step by step solution:
Given
$\csc \left( {\dfrac{{11\pi }}{6}} \right)...................................\left( i \right)$
Now we also know$\csc x = \dfrac{1}{{\sin x}}$. So to find the value of $\csc \left( {\dfrac{{11\pi }}{6}}
\right)$we need to find $\sin \left( {\dfrac{{11\pi }}{6}} \right)$and then find it’s reciprocal.
Now to find $\sin \left( {\dfrac{{11\pi }}{6}} \right)$we can use the identity$\;\cos 2\theta = 1 - 2{\sin
^2}\theta $.
Finding the value of$\sin \left( {\dfrac{{11\pi }}{6}} \right)$:
Now let’s assume $\sin \left( {\dfrac{{11\pi }}{6}} \right) = \sin a......................\left( {ii} \right)$
So similarly we can write $\;\cos a = \cos \left( {\dfrac{{11\pi }}{6}} \right)$
$ \Rightarrow \cos 2a = \cos \left( {\dfrac{{22\pi }}{6}} \right)$
We have to find the value of $\cos \left( {\dfrac{{22\pi }}{6}} \right)$such that by using the identity we can then solve the question.
So finding the value of$\cos \left( {\dfrac{{22\pi }}{6}} \right)$:
We know that $\cos \left( {\dfrac{{22\pi }}{6}} \right)$can be written as
$
\cos \left( {\dfrac{{12\left( {2\pi } \right)}}{6} - \dfrac{{2\pi }}{6}} \right) = \cos \left( {2\left( {2\pi }
\right) - \dfrac{{2\pi }}{6}} \right).................(iii) \\
\\
$
So from (iii) we know that \[\cos \left( {2\left( {2\pi } \right) - \dfrac{{2\pi }}{6}} \right)\]would be in
the IV Quadrant where cosine is positive.
Such that:
\[\cos \left( {2\left( {2\pi } \right) - \dfrac{{2\pi }}{6}} \right) = \cos \left( {\dfrac{{2\pi }}{6}}
\right)..................(iv)\]
Also we know \[\cos \left( {\dfrac{{2\pi }}{6}} \right) = \cos \left( {\dfrac{\pi }{3}} \right) =
\dfrac{1}{2}....................(v)\]
Now by using the identity $\;\cos 2\theta = 1 - 2{\sin ^2}\theta $we get
$
\Rightarrow \cos 2a = 1 - 2{\sin ^2}a = \dfrac{1}{2} \\
\Rightarrow 2{\sin ^2}a = 1 - \dfrac{1}{2} = \dfrac{1}{2} \\
\Rightarrow {\sin ^2}a = \dfrac{1}{4} \\
$
Now from (ii) \[\sin \left( {\dfrac{{11\pi }}{6}} \right) = \sin a\]
$
\Rightarrow \sin {\left( {\dfrac{{11\pi }}{6}} \right)^2} = \dfrac{1}{4} \\
\Rightarrow \sin \left( {\dfrac{{11\pi }}{6}} \right) = - \dfrac{1}{2} \\
$
Now we know that$\csc x = \dfrac{1}{{\sin x}}$, such that:
$\csc \left( {\dfrac{{11\pi }}{6}} \right) = \dfrac{1}{{\sin \left( {\dfrac{{11\pi }}{6}} \right)}} =
\dfrac{1}{{\left( { - \dfrac{1}{2}} \right)}} = - 2$
Therefore the value of$\csc \left( {\dfrac{{11\pi }}{6}} \right)\;{\text{is}}\; - 2$.
Note:
General things to be known for solving this question.
I Quadrant:$0\; - \;\dfrac{\pi }{2}$ All values are positive.
II Quadrant:$\dfrac{\pi }{2}\; - \;\pi $ Only Sine and Cosec values are positive.
III Quadrant:$\pi \; - \;\dfrac{{3\pi }}{2}$ Only Tan and Cot values are positive.
IV Quadrant:$\dfrac{{3\pi }}{2}\; - \;2\pi $ Only Cos and Sec values are positive.
Some other equations needed for solving these types of problem are:
\[
\sin \left( {2\theta } \right) = 2\sin \left( \theta \right)\cos \left( \theta \right) \\
\cos \left( {2\theta } \right) = {\cos ^2}\left( \theta \right)-{\sin ^2}\left( \theta \right) = 1-2{\text{
}}{\sin ^2}\left( \theta \right) = 2{\text{ }}{\cos ^2}\left( \theta \right)-1 \\
\]
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)