Question

If $\sin \left(\theta \right)+\mathrm{cosec}\left(\theta \right)=2$, then ${\sin }^2\left(\theta \right)+{\mathrm{cosec}}^2\left(\theta \right)=$

• A. $1$
• B. $4$
• C. $2$
• D. None of these

Answer 1

The correct option is C

$2$

Explanation for the correct option:

Step 1: Formation of the quadratic equation

Here, in the given equation we can write $\mathrm{cosec}\left(\theta \right)$ as $\frac{1}{\sin \left(\theta \right)}$,

$\begin{array}{rcl}\therefore \sin \left(\theta \right)+\frac{1}{\sin \left(\theta \right)}& =& 2\\ & \Rightarrow & {\sin }^2\left(\theta \right)+1=2\sin \left(\theta \right)\\ & \Rightarrow & {\sin }^2\left(\theta \right)-2\sin \left(\theta \right)+1=0\end{array}$

Step 2: Determination of the value of the required expression

Solving the above equation to determine the value of $\begin{array}{rcl}& & \sin \left(\theta \right)\end{array}$ by splitting the middle term,

$\begin{array}{rcl}\therefore {\sin }^2\left(\theta \right)-2\sin \left(\theta \right)+1& =& 0\\ & \Rightarrow & {\sin }^2\left(\theta \right)-\sin \left(\theta \right)-\sin \left(\theta \right)+1=0\\ & \Rightarrow & \sin \left(\theta \right)\left(\sin \left(\theta \right)-1\right)-1\left(\sin \left(\theta \right)-1\right)=0\\ & \Rightarrow & \left(\sin \left(\theta \right)-1\right)\left(\sin \left(\theta \right)-1\right)=0\\ & \Rightarrow & {\left(\sin \left(\theta \right)-1\right)}^2=0\\ & \Rightarrow & \sin \left(\theta \right)-1=0\\ & \Rightarrow & \sin \left(\theta \right)=1\end{array}$

Now, from the given equation we have

$\mathrm{cosec}\left(\theta \right)=1$

Then, the required equation will be,

$\begin{array}{rcl}\therefore {\sin }^2\left(\theta \right)+{\mathrm{cosec}}^2\left(\theta \right)& =& {\left(1\right)}^2+{\left(1\right)}^2\\ & =& 2\end{array}$

Hence, the correct option is (C).