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Question: If \(\sin x+\text{cosec }x=2\) , then \({{\sin }^{n}}x+\text{cose}{{\text{c}}^{n}}\text{ }x\) is equ...

If sin⁡x+cosec x=2\sin x+\text{cosec }x=2 , then sin⁡nx+cosecn x{{\sin }^{n}}x+\text{cose}{{\text{c}}^{n}}\text{ }x is equal to
(a) 22
(b) 2n{{2}^{n}}
(c) 2n−1{{2}^{n-1}}
(d) 2n−2{{2}^{n-2}}

Explanation

Solution

We can substitute cosec x=1sin⁡x\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\sin x . We now need to substitute this value in the equation sin⁡nx+cosecn x{{\sin }^{n}}x+\text{cose}{{\text{c}}^{n}}\text{ }x after replacing the trigonometric ratio cosec x=1sin⁡x\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 cosec x=1sin⁡x\text{cosec }x=\dfrac{1}{\sin x} . Using this formula, we can write
sin⁡x+1sin⁡x=2\sin x+\dfrac{1}{\sin x}=2
We can simplify by taking LCM,
sin⁡2x+1sin⁡x=2\dfrac{{{\sin }^{2}}x+1}{\sin x}=2
Rearranging the terms, we get
sin⁡2x+1=2sin⁡x{{\sin }^{2}}x+1=2\sin x
Or, we can write,
sin⁡2x−2sin⁡x+1=0{{\sin }^{2}}x-2\sin x+1=0
Let us substitute m=sin⁡xm=\sin x .
So, we now have the quadratic equation,
m2−2m+1=0{{m}^{^{2}}}-2m+1=0
We know that (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} . Using this equation, we get
(m−1)2=0{{\left( m-1 \right)}^{2}}=0
Taking square root on both sides, we get
m−1=0m-1=0
Or, m=1m=1
Thus, we now have, sin⁡x=1\sin x=1 .
So, we can easily write
sin⁡nx=(1)n{{\sin }^{n}}x={{\left( 1 \right)}^{n}}
So, we get sin⁡nx=1...(i){{\sin }^{n}}x=1...\left( i \right)
Now, since sin⁡x=1\sin x=1 , we can also write that
1sin⁡x=1\dfrac{1}{\sin x}=1
which is the same as cosec x=1\text{cosec }x=1 .
Hence, we can also write, cosecn x=(1)n\text{cose}{{\text{c}}^{n}}\text{ }x={{\left( 1 \right)}^{n}}
Thus, cosecn x=1...(ii)\text{cose}{{\text{c}}^{n}}\text{ }x=1...\left( ii \right)
We can now add the equations (i) and (ii) to get
sin⁡nx+cosecn x=1+1{{\sin }^{n}}x+\text{cose}{{\text{c}}^{n}}\text{ }x=1+1
Hence, we have the required result as
sin⁡nx+cosecn x=2.{{\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’.