Answer
405.6k+ views
Hint: In this question, from the given values of cosec function by using the trigonometric identities we can find the values of the sine, cosine and cot functions. Then on substituting the respective values in the given expression of the question we can calculate the right hand side and left hand side values. Then on comparing the values obtained we get the result.
\[\begin{align}
& \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
& \text{cose}{{\text{c}}^{2}}\theta -1={{\cot }^{2}}\theta \\
& \cos \theta =\cot \theta \cdot \sin \theta \\
\end{align}\]
Complete step by step answer:
Now, from the given question we have
\[\text{cosec}\theta =2\]
Now, by using the trigonometric identity which gives the relation between cosec function and sine function we get,
\[\Rightarrow \text{cosec}\theta =\dfrac{1}{\sin \theta }\]
Now, this can also be written as
\[\Rightarrow \sin \theta =\dfrac{1}{\text{cosec}\theta }\]
Let us now substitute the value of secant in this
\[\Rightarrow \sin \theta =\dfrac{1}{2}\]
Now, using the relation between the cosec function and cot function from the trigonometric identities we have
\[\Rightarrow {{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta =1\]
Now, this can also be written as
\[\Rightarrow {{\cot }^{2}}\theta ={{\operatorname{cosec}}^{2}}\theta -1\]
Now, on substituting the value of cosec we get,
\[\Rightarrow {{\cot }^{2}}\theta ={{\left( 2 \right)}^{2}}-1\]
Now, on further simplification we have
\[\Rightarrow {{\cot }^{2}}\theta =4-1=3\]
Let us now apply the square root on both the sides
\[\Rightarrow \sqrt{{{\cot }^{2}}\theta }=\sqrt{3}\]
Now, on simplifying it we get,
\[\therefore \cot \theta =\sqrt{3}\]
Now, by using the relation between sine, cosine and cot functions from the trigonometric identities we have,
\[\Rightarrow \dfrac{\cos \theta }{\sin \theta }=\cot \theta \]
Now, this can also be written as
\[\Rightarrow \cos \theta =\cot \theta \times \sin \theta \]
Now, on substituting the respective values in the above equation we get,
\[\Rightarrow \cos \theta =\sqrt{3}\times \dfrac{1}{2}\]
Now, on simplifying further we get,
\[\therefore \cos \theta =\dfrac{\sqrt{3}}{2}\]
Now, from the given expression in the question we have
\[\begin{align}
& L.H.S=\left( \sqrt{3}+\dfrac{\dfrac{1}{2}}{1+\dfrac{\sqrt{3}}{2}} \right) \\
&\Rightarrow L.H.S=\left( \sqrt{3}+\dfrac{\dfrac{1}{2}}{\dfrac{2+\sqrt{3}}{2}} \right) \\
\end{align}\]
In the above equation, $\dfrac{1}{2}$ will be cancelled from the numerator and denominator and we are left with:
\[L.H.S=\left( \sqrt{3}+\dfrac{1}{2+\sqrt{3}} \right)\]
Taking $2+\sqrt{3}$ as L.C.M then we get,
\[\begin{align}
& L.H.S=\left( \dfrac{2\sqrt{3}+3+1}{2+\sqrt{3}} \right) \\
&\Rightarrow L.H.S=\left( \dfrac{2\sqrt{3}+4}{2+\sqrt{3}} \right) \\
\end{align}\]
We can take 2 as common from the numerator.
\[\begin{align}
& L.H.S=2\left( \dfrac{\sqrt{3}+2}{2+\sqrt{3}} \right) \\
&\Rightarrow L.H.S=2 \\
\end{align}\]
Thus, the value of right hand side is equal to left hand side
Hence, it is verified that \[\left( \cot \theta +\dfrac{\sin \theta }{1+\cos \theta } \right)=2\]
Note: The other way of solving the above problem is that we have given:
$\text{cosec}\theta =2$
We know that $\text{cosec}\theta =\dfrac{1}{\sin \theta }$ so using this relation in the above we get,
$\begin{align}
& \dfrac{1}{\sin \theta }=2 \\
& \Rightarrow \sin \theta =\dfrac{1}{2} \\
\end{align}$
And we know that, the value of $\sin \theta =\dfrac{1}{2}$ when $\theta ={{30}^{\circ }}$ so to verify the following relation we can put $\theta ={{30}^{\circ }}$ and then prove it.
\[\left( \cot \theta +\dfrac{\sin \theta }{1+\cos \theta } \right)=2\]
$\Rightarrow \cot {{30}^{\circ }}+\dfrac{\sin {{30}^{\circ }}}{1+\cos {{30}^{\circ }}}=2$ ……… Eq. (1)
We know the value of following trigonometric ratios as follows:
$\begin{align}
& \cot {{30}^{\circ }}=\sqrt{3} \\
& \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
\end{align}$
Substituting the above values in eq. (1) we get,
$\begin{align}
& \sqrt{3}+\dfrac{\dfrac{1}{2}}{1+\dfrac{\sqrt{3}}{2}}=2 \\
& \Rightarrow \sqrt{3}+\dfrac{\dfrac{1}{2}}{\dfrac{2+\sqrt{3}}{2}}=2 \\
\end{align}$
In the above equation, $\dfrac{1}{2}$ will be cancelled out from the numerator and denominator and we get,
$\sqrt{3}+\dfrac{1}{2+\sqrt{3}}=2$
Now, rationalizing $\dfrac{1}{2+\sqrt{3}}$ by multiplying and dividing $2-\sqrt{3}$ with this dfraction we get,
$\sqrt{3}+\dfrac{1}{2+\sqrt{3}}\times \left( \dfrac{2-\sqrt{3}}{2-\sqrt{3}} \right)=2$
In the denominator we have $\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)$ which is written in the form of $\left( a+b \right)\left( a-b \right)$ and we know that:
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ so using this relation in the above equation we get,
$\begin{align}
& \Rightarrow \sqrt{3}+\dfrac{2-\sqrt{3}}{{{2}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}=2 \\
& \Rightarrow \sqrt{3}+\dfrac{2-\sqrt{3}}{4-3}=2 \\
& \Rightarrow \sqrt{3}+2-\sqrt{3}=2 \\
& \Rightarrow 2=2 \\
\end{align}$
As you can see that L.H.S is equal to R.H.S so we have proved the given equation.
\[\begin{align}
& \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
& \text{cose}{{\text{c}}^{2}}\theta -1={{\cot }^{2}}\theta \\
& \cos \theta =\cot \theta \cdot \sin \theta \\
\end{align}\]
Complete step by step answer:
Now, from the given question we have
\[\text{cosec}\theta =2\]
Now, by using the trigonometric identity which gives the relation between cosec function and sine function we get,
\[\Rightarrow \text{cosec}\theta =\dfrac{1}{\sin \theta }\]
Now, this can also be written as
\[\Rightarrow \sin \theta =\dfrac{1}{\text{cosec}\theta }\]
Let us now substitute the value of secant in this
\[\Rightarrow \sin \theta =\dfrac{1}{2}\]
Now, using the relation between the cosec function and cot function from the trigonometric identities we have
\[\Rightarrow {{\operatorname{cosec}}^{2}}\theta -{{\cot }^{2}}\theta =1\]
Now, this can also be written as
\[\Rightarrow {{\cot }^{2}}\theta ={{\operatorname{cosec}}^{2}}\theta -1\]
Now, on substituting the value of cosec we get,
\[\Rightarrow {{\cot }^{2}}\theta ={{\left( 2 \right)}^{2}}-1\]
Now, on further simplification we have
\[\Rightarrow {{\cot }^{2}}\theta =4-1=3\]
Let us now apply the square root on both the sides
\[\Rightarrow \sqrt{{{\cot }^{2}}\theta }=\sqrt{3}\]
Now, on simplifying it we get,
\[\therefore \cot \theta =\sqrt{3}\]
Now, by using the relation between sine, cosine and cot functions from the trigonometric identities we have,
\[\Rightarrow \dfrac{\cos \theta }{\sin \theta }=\cot \theta \]
Now, this can also be written as
\[\Rightarrow \cos \theta =\cot \theta \times \sin \theta \]
Now, on substituting the respective values in the above equation we get,
\[\Rightarrow \cos \theta =\sqrt{3}\times \dfrac{1}{2}\]
Now, on simplifying further we get,
\[\therefore \cos \theta =\dfrac{\sqrt{3}}{2}\]
Now, from the given expression in the question we have
\[\begin{align}
& L.H.S=\left( \sqrt{3}+\dfrac{\dfrac{1}{2}}{1+\dfrac{\sqrt{3}}{2}} \right) \\
&\Rightarrow L.H.S=\left( \sqrt{3}+\dfrac{\dfrac{1}{2}}{\dfrac{2+\sqrt{3}}{2}} \right) \\
\end{align}\]
In the above equation, $\dfrac{1}{2}$ will be cancelled from the numerator and denominator and we are left with:
\[L.H.S=\left( \sqrt{3}+\dfrac{1}{2+\sqrt{3}} \right)\]
Taking $2+\sqrt{3}$ as L.C.M then we get,
\[\begin{align}
& L.H.S=\left( \dfrac{2\sqrt{3}+3+1}{2+\sqrt{3}} \right) \\
&\Rightarrow L.H.S=\left( \dfrac{2\sqrt{3}+4}{2+\sqrt{3}} \right) \\
\end{align}\]
We can take 2 as common from the numerator.
\[\begin{align}
& L.H.S=2\left( \dfrac{\sqrt{3}+2}{2+\sqrt{3}} \right) \\
&\Rightarrow L.H.S=2 \\
\end{align}\]
Thus, the value of right hand side is equal to left hand side
Hence, it is verified that \[\left( \cot \theta +\dfrac{\sin \theta }{1+\cos \theta } \right)=2\]
Note: The other way of solving the above problem is that we have given:
$\text{cosec}\theta =2$
We know that $\text{cosec}\theta =\dfrac{1}{\sin \theta }$ so using this relation in the above we get,
$\begin{align}
& \dfrac{1}{\sin \theta }=2 \\
& \Rightarrow \sin \theta =\dfrac{1}{2} \\
\end{align}$
And we know that, the value of $\sin \theta =\dfrac{1}{2}$ when $\theta ={{30}^{\circ }}$ so to verify the following relation we can put $\theta ={{30}^{\circ }}$ and then prove it.
\[\left( \cot \theta +\dfrac{\sin \theta }{1+\cos \theta } \right)=2\]
$\Rightarrow \cot {{30}^{\circ }}+\dfrac{\sin {{30}^{\circ }}}{1+\cos {{30}^{\circ }}}=2$ ……… Eq. (1)
We know the value of following trigonometric ratios as follows:
$\begin{align}
& \cot {{30}^{\circ }}=\sqrt{3} \\
& \cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2} \\
\end{align}$
Substituting the above values in eq. (1) we get,
$\begin{align}
& \sqrt{3}+\dfrac{\dfrac{1}{2}}{1+\dfrac{\sqrt{3}}{2}}=2 \\
& \Rightarrow \sqrt{3}+\dfrac{\dfrac{1}{2}}{\dfrac{2+\sqrt{3}}{2}}=2 \\
\end{align}$
In the above equation, $\dfrac{1}{2}$ will be cancelled out from the numerator and denominator and we get,
$\sqrt{3}+\dfrac{1}{2+\sqrt{3}}=2$
Now, rationalizing $\dfrac{1}{2+\sqrt{3}}$ by multiplying and dividing $2-\sqrt{3}$ with this dfraction we get,
$\sqrt{3}+\dfrac{1}{2+\sqrt{3}}\times \left( \dfrac{2-\sqrt{3}}{2-\sqrt{3}} \right)=2$
In the denominator we have $\left( 2+\sqrt{3} \right)\left( 2-\sqrt{3} \right)$ which is written in the form of $\left( a+b \right)\left( a-b \right)$ and we know that:
$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$ so using this relation in the above equation we get,
$\begin{align}
& \Rightarrow \sqrt{3}+\dfrac{2-\sqrt{3}}{{{2}^{2}}-{{\left( \sqrt{3} \right)}^{2}}}=2 \\
& \Rightarrow \sqrt{3}+\dfrac{2-\sqrt{3}}{4-3}=2 \\
& \Rightarrow \sqrt{3}+2-\sqrt{3}=2 \\
& \Rightarrow 2=2 \\
\end{align}$
As you can see that L.H.S is equal to R.H.S so we have proved the given equation.
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)