Question

If $\sin \left(\theta \right)+\mathrm{cosec}\left(\theta \right)=2$, then the value of ${\sin }^{10}\left(\theta \right)+{\mathrm{cosec}}^{10}\left(\theta \right)$ is

• A. $2$
• B. $210$
• C. $29$
• D. $10$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

Step 1: Formation of the quadratic equation

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 $\mathbf{sin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}\mathbf{,}\mathbf{}\mathbf{cosec}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}$

Solve the above quadratic equation by splitting the middle term,

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

This gives us the value of $\sin \left(\theta \right)$ that is $1$.

Now, $\mathrm{cosec}\left(\theta \right)$ will be

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

Step 3: Calculation for the required expression

Since $\sin \left(\theta \right)=1\text{and}\mathrm{cosec}\left(\theta \right)=1$, then

$\begin{array}{rcl}{\sin }^{10}\left(\theta \right)+{\mathrm{cosec}}^{10}\left(\theta \right)& =& {\left(1\right)}^{10}+{\left(1\right)}^{10}\\ & =& 1+1\\ & =& 2\end{array}$

Hence, the correct option is (A).