
If \[\text{cosec}\theta =2\], verify that:
\[\left( \cot \theta +\dfrac{\sin \theta }{1+\cos \theta } \right)=2\]
Answer
465k+ 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
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
Define least count of vernier callipers How do you class 11 physics CBSE

The combining capacity of an element is known as i class 11 chemistry CBSE

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Find the image of the point 38 about the line x+3y class 11 maths CBSE

Can anyone list 10 advantages and disadvantages of friction

Distinguish between Mitosis and Meiosis class 11 biology CBSE
