Question

If $\sin \theta +\operatorname{cosec}\theta =2$, then find the value of ${{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta $?

Answer 1

Hint: We start solving the problem by squaring the both sides of the given trigonometric equation $\sin \theta +\operatorname{cosec}\theta =2$. We then use expand ${{\left( \sin \theta +\operatorname{cosec}\theta \right)}^{2}}$ using the fact that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. We then use the property of trigonometric functions that $\sin \theta \times \operatorname{cosec}\theta =1$ and make the necessary calculations to get the required value of ${{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta $.

Complete step by step answer:
According to the problem, we are given that the sum of the trigonometric functions as $\sin \theta +\operatorname{cosec}\theta =2$ and we need to find the value of ${{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta $.
Now, we have $\sin \theta +\operatorname{cosec}\theta =2$ ---(1).
Let us do squaring on both sides in equation (1).
$\Rightarrow {{\left( \sin \theta +\operatorname{cosec}\theta \right)}^{2}}={{2}^{2}}$ ---(2).
We know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. We use this result in equation (2)
$\Rightarrow {{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta +2\left( \sin \theta \right)\left( \operatorname{cosec}\theta \right)=4$ ---(3).
We know that $\sin \theta \times \operatorname{cosec}\theta =1$. We use this result in equation (3).
$\Rightarrow {{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta +2\times 1=4$.
$\Rightarrow {{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta +2=4$.
$\Rightarrow {{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta =2$.

∴ We have found the value of ${{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta $ as 2.

Note: We can also solve this problem as shown below:
We have $\sin \theta +\operatorname{cosec}\theta =2$ ---(4).
We know that $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$. We use this result in equation (4).
$\Rightarrow \sin \theta +\dfrac{1}{\sin \theta }=2$.
$\Rightarrow \dfrac{{{\sin }^{2}}\theta +1}{\sin \theta }=2$.
$\Rightarrow {{\sin }^{2}}\theta +1=2\sin \theta $.
$\Rightarrow {{\sin }^{2}}\theta -2\sin \theta +1=0$.
$\Rightarrow {{\left( \sin \theta -1 \right)}^{2}}=0$.
\[\Rightarrow \sin \theta -1=0\].
\[\Rightarrow \sin \theta =1\].
Let us now find the value of $\operatorname{cosec}\theta $.
So, we have $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$.
$\Rightarrow \operatorname{cosec}\theta =\dfrac{1}{1}=1$.
Now, let us find the value of ${{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta $.
So, we have \[{{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta ={{1}^{2}}+{{1}^{2}}\].
$\Rightarrow {{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta =1+1$.
$\Rightarrow {{\sin }^{2}}\theta +{{\operatorname{cosec}}^{2}}\theta =2$.
We can also solve this problem by performing trial and error method for the angle $\theta $ in the $\sin \theta +\operatorname{cosec}\theta =2$. Similarly, we can expect problems to find the value of ${{\sin }^{n}}\theta +{{\operatorname{cosec}}^{n}}\theta $ using the obtained value of $\sin \theta $ and $\operatorname{cosec}\theta $.