Question

Find the value of inverse trigonometric function ${\cot ^{ - 1}}\left[ {\dfrac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}} \right]$ where $0 < x < \dfrac{\pi }{2}$.
$
  {\text{A}}{\text{. }}\dfrac{x}{2} \\
  {\text{B}}{\text{. }}\dfrac{\pi }{2} - 2x \\
  {\text{C}}{\text{. }}2\pi - x \\
  {\text{D}}{\text{. }}\pi - \dfrac{x}{2} \\
 $

Answer 1

Hint: Here, we will convert the term inside the inverse cotangent function i.e., $\dfrac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}$ in terms of cotangent function of some angle and then we will use the formula ${\cot ^{ - 1}}\left[ {\cot \theta } \right] = \theta $ in order to obtain the value of the required expression.

Complete step-by-step answer:

Let us suppose the function $y = \dfrac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}{\text{ }} \to {\text{(1)}}$.
In order to rationalize the function given in the RHS of equation (1), we will multiply and divide the RHS with the term$\left[ {\sqrt {1 - \sin x} - \sqrt {1 + \sin x} } \right]$. Now, equation (1) becomes
\[
  y = \left[ {\dfrac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}} \right] \times \left[ {\dfrac{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}} \right] \\
   \Rightarrow y = \left[ {\dfrac{{\left( {\sqrt {1 - \sin x} + \sqrt {1 + \sin x} } \right)\left( {\sqrt {1 - \sin x} - \sqrt {1 + \sin x} } \right)}}{{\left( {\sqrt {1 - \sin x} - \sqrt {1 + \sin x} } \right)\left( {\sqrt {1 - \sin x} - \sqrt {1 + \sin x} } \right)}}} \right] \\
   \Rightarrow y = \left[ {\dfrac{{{{\left( {\sqrt {1 - \sin x} } \right)}^2} - {{\left( {\sqrt {1 + \sin x} } \right)}^2}}}{{{{\left( {\sqrt {1 - \sin x} - \sqrt {1 + \sin x} } \right)}^2}}}} \right] \\
   \Rightarrow y = \left[ {\dfrac{{1 - \sin x - \left( {1 + \sin x} \right)}}{{{{\left( {\sqrt {1 - \sin x} } \right)}^2} + {{\left( {\sqrt {1 + \sin x} } \right)}^2} - 2\left( {\sqrt {1 - \sin x} } \right)\left( {\sqrt {1 + \sin x} } \right)}}} \right] \\
   \Rightarrow y = \left[ {\dfrac{{1 - \sin x - 1 - \sin x}}{{1 - \sin x + 1 + \sin x - 2\left[ {\sqrt {\left( {1 - \sin x} \right)\left( {1 + \sin x} \right)} } \right]}}} \right] \\
   \Rightarrow y = \left[ {\dfrac{{ - 2\sin x}}{{2 - 2\left[ {\sqrt {1 - \sin x + \sin x - {{\left( {\sin x} \right)}^2}} } \right]}}} \right] \\
   \Rightarrow y = \left[ {\dfrac{{ - 2\sin x}}{{2\left[ {1 - \sqrt {1 - {{\left( {\sin x} \right)}^2}} } \right]}}} \right]{\text{ }} \to {\text{(1)}} \\
 \]
As we know that ${\left( {\sin x} \right)^2} + {\left( {\cos x} \right)^2} = 1$
$
   \Rightarrow {\left( {\cos x} \right)^2} = 1 - {\left( {\sin x} \right)^2} \\
   \Rightarrow \cos x = \sqrt {1 - {{\left( {\sin x} \right)}^2}} {\text{ }} \to {\text{(2)}} \\
 $
By substituting the equation (2) in equation (1), we get
\[
   \Rightarrow y = \left[ {\dfrac{{ - 2\sin x}}{{2\left[ {1 - \cos x} \right]}}} \right] \\
   \Rightarrow y = \left[ {\dfrac{{ - \sin x}}{{1 - \cos x}}} \right]{\text{ }} \to (3) \\
 \]
Also, we know that $\sin x = 2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right){\text{ }} \to {\text{(4)}}$ and $
  \cos x = 1 - 2{\left[ {\sin \left( {\dfrac{x}{2}} \right)} \right]^2} \\
   \Rightarrow 2{\left[ {\sin \left( {\dfrac{x}{2}} \right)} \right]^2} = 1 - \cos x{\text{ }} \to {\text{(5)}} \\
 $
By substituting the equations (4) and (5) in equation (3), we get
\[
   \Rightarrow y = \left[ {\dfrac{{ - 2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)}}{{2{{\left[ {\sin \left( {\dfrac{x}{2}} \right)} \right]}^2}}}} \right] \\
   \Rightarrow y = \left[ {\dfrac{{ - \cos \left( {\dfrac{x}{2}} \right)}}{{\sin \left( {\dfrac{x}{2}} \right)}}} \right] \\
   \Rightarrow y = - \cot \left( {\dfrac{x}{2}} \right){\text{ }} \to {\text{(6)}} \\
 \]
Also, $
  \cot \left( {\pi - \theta } \right) = - \cot \theta \\
   \Rightarrow \cot \left( {\pi - \dfrac{x}{2}} \right) = - \cot \left( {\dfrac{x}{2}} \right){\text{ }} \to {\text{(7)}} \\
 $
Using the equation (7) in equation (6), we get
$ \Rightarrow y = \cot \left( {\pi - \dfrac{x}{2}} \right){\text{ }} \to {\text{(8)}}$
According to the equations (1) and (8), we get
$y = \dfrac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }} = \cot \left( {\pi - \dfrac{x}{2}} \right){\text{ }} \to {\text{(9)}}$
Using equation (9) in order to obtain the value of the expression needed, we get
${\cot ^{ - 1}}\left[ {\dfrac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}} \right] = {\cot ^{ - 1}}\left[ {\cot \left( {\pi - \dfrac{x}{2}} \right)} \right]$
By using the formula i.e., ${\cot ^{ - 1}}\left[ {\cot \theta } \right] = \theta $, above equation becomes
${\cot ^{ - 1}}\left[ {\dfrac{{\sqrt {1 - \sin x} + \sqrt {1 + \sin x} }}{{\sqrt {1 - \sin x} - \sqrt {1 + \sin x} }}} \right] = \left( {\pi - \dfrac{x}{2}} \right)$
Hence, option D is correct.

Note: In this particular problem, we have used the formula $\sin 2\theta = 2\sin \theta \cos \theta $ and replaced angle $\theta $ with $\dfrac{x}{2}$ in order to get $\sin x = 2\sin \left( {\dfrac{x}{2}} \right)\cos \left( {\dfrac{x}{2}} \right)$ and also we have use the formula $\cos 2\theta = 1 - 2{\left[ {\sin \theta } \right]^2}$ and replaced angle $\theta $ with $\dfrac{x}{2}$ in order to get $\cos x = 1 - 2{\left[ {\sin \left( {\dfrac{x}{2}} \right)} \right]^2}$.