Question

If $\sin x+\text{cosec }x=2$ , then ${{\sin }^{n}}x+\text{cose}{{\text{c}}^{n}}\text{ }x$ is equal to
(a) $2$
(b) ${{2}^{n}}$
(c) ${{2}^{n-1}}$
(d) ${{2}^{n-2}}$

Answer 1

Hint: We can substitute $\text{cosec }x=\dfrac{1}{\sin x}$ and by doing so, we will get a quadratic equation. We can easily solve this quadratic equation to find the value of $\sin x$ . We now need to substitute this value in the equation ${{\sin }^{n}}x+\text{cose}{{\text{c}}^{n}}\text{ }x$ after replacing the trigonometric ratio $\text{cosec }x=\dfrac{1}{\sin x}$ . We will thus get the required result.

Complete step by step answer:
From the definition of trigonometric ratios, we know that $\text{cosec }x=\dfrac{1}{\sin x}$ . Using this formula, we can write
$\sin x+\dfrac{1}{\sin x}=2$
We can simplify by taking LCM,
$\dfrac{{{\sin }^{2}}x+1}{\sin x}=2$
Rearranging the terms, we get
${{\sin }^{2}}x+1=2\sin x$
Or, we can write,
${{\sin }^{2}}x-2\sin x+1=0$
Let us substitute $m=\sin x$ .
So, we now have the quadratic equation,
${{m}^{^{2}}}-2m+1=0$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . Using this equation, we get
${{\left( m-1 \right)}^{2}}=0$
Taking square root on both sides, we get
$m-1=0$
Or, $m=1$
Thus, we now have, $\sin x=1$ .
So, we can easily write
${{\sin }^{n}}x={{\left( 1 \right)}^{n}}$
So, we get ${{\sin }^{n}}x=1...\left( i \right)$
Now, since $\sin x=1$ , we can also write that
$\dfrac{1}{\sin x}=1$
which is the same as $\text{cosec }x=1$ .
Hence, we can also write, $\text{cose}{{\text{c}}^{n}}\text{ }x={{\left( 1 \right)}^{n}}$
Thus, $\text{cose}{{\text{c}}^{n}}\text{ }x=1...\left( ii \right)$
We can now add the equations (i) and (ii) to get
${{\sin }^{n}}x+\text{cose}{{\text{c}}^{n}}\text{ }x=1+1$
Hence, we have the required result as
${{\sin }^{n}}x+\text{cose}{{\text{c}}^{n}}\text{ }x=2.$

So, the correct answer is “Option a”.

Note: To solve this problem faster and efficiently for an objective paper, we can substitute a few values for n and find the suitable option. By putting n = 2 and n = 3, we can get the result efficiently. We must note that this is just an objective approach, and can be used only when there is no option like ‘None of these’.