Question

If \[\sin x+\text{cosec} x=2\], then \[{{\sin }^{n}}x+cose{{c}^{n}}x=?\]
1 \[2\]
2 \[{{2}^{n}}\]
3 \[{{2}^{n-1}}\]
4 \[{{2}^{n-2}}\]

Answer 1

Hint: In order to find the value of \[{{\sin }^{n}}x+cose{{c}^{n}}x\] when \[\sin x+\text{cosec} x=2\] , firstly we will be converting the given \[\text{cosec} x\] in terms of \[\sin x\]. Then we will be taking the LCM of the functions and solving them. Then we will be transposing the terms for our easy calculation and form an equation. Upon solving it accordingly, we will be obtaining our required solution.

Complete step-by-step solution:
Let us have a brief regarding the trigonometric functions. The counter-clockwise angle between the initial arm and the terminal arm of an angle in standard position is called the principal angle. Its value is between \[{{0}^{\circ }}\] and \[{{360}^{\circ }}\]. The relationship between the angles and sides of a triangle are given by the trigonometric functions. The basic trigonometric functions are sine, cosine, tangent, cotangent, secant and cosecant. These are the basic main trigonometric functions used.
Now let start finding the value of \[{{\sin }^{n}}x+cose{{c}^{n}}x\] when \[\sin x+\text{cosec} x=2\]
We are given that \[\sin x+\text{cosec} x=2\]
Firstly, we will be converting the \[\text{cosec} x\] in terms of \[\sin x\]. We get,
\[\sin x+\dfrac{1}{\sin x}=2\]
Now, upon taking LCM and solving, we get
\[\Rightarrow \dfrac{{{\sin }^{2}}x+1}{\sin x}=2\]
Now we will transpose the \[\sin x\] from the denominator to the RHS. We get,
\[\Rightarrow {{\sin }^{2}}x+1=2\sin x\]
Upon solving this,
\[\Rightarrow {{\sin }^{2}}x+1-2\sin x=0\]
Now we can observe that the obtained equation is in the terms of identity \[{{\left( a-b \right)}^{2}}\]. So upon writing it in terms of \[{{\left( a-b \right)}^{2}}\], we get
\[\Rightarrow {{\left( \sin x-1 \right)}^{2}}=0\]
Upon solving this,
\[\begin{align}
  & \Rightarrow {{\left( \sin x-1 \right)}^{2}}=0 \\
 & \Rightarrow \sin x-1=0 \\
 & \Rightarrow \sin x=1 \\
\end{align}\]
\[\because \text{cosec} x=\dfrac{1}{\sin x}=\dfrac{1}{1}=1\]
So we now have the required values. Upon substituting it, we get
\[{{\sin }^{n}}x+cose{{c}^{n}}x={{1}^{n}}+{{1}^{n}}=2\]
\[\therefore \]Option 1 is the correct answer.

Note: We must always have a note that one power anything would result as one itself. We must always try to expand it into the identities whose values we are aware of, for easy calculations just like we have used \[{{\left( a-b \right)}^{2}}\] as an identity to be expressed.