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# If $\text{cosec}\theta =2$, verify that:$\left( \cot \theta +\dfrac{\sin \theta }{1+\cos \theta } \right)=2$

Last updated date: 13th Jun 2024
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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}

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.