Question

If \[y={{\cot }^{-1}}\left[ \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right]\] where \[x\in \left( 0,\dfrac{\pi }{4} \right)\] then find the value of \[\dfrac{dy}{dx}\] .
(a) \[\dfrac{1}{2}\]
(b) \[\dfrac{2}{3}\]
(c) 3
(d) 1

Answer 1

Hint: We will be using the concepts of inverse trigonometric function to simplify the expression we will simplify the value of \[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\] first so that it become easy to solve the question further , for simplifying the expression we will be using trigonometric identities like
\[\begin{align}
  & \sin 2x=2\sin x\cos x \\
 & \cos 2x=2{{\cos }^{2}}x-1 \\
 & {{\cot }^{-1}}\left( \cot x \right)=x \\
\end{align}\]
 and then we use the concepts of differential calculus to find the final answer.

Complete step-by-step answer:
Now, \[y={{\cot }^{-1}}\left[ \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right]\]
Now, we will first simplify \[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\]
We will now multiply both numerator and denominator by \[\sqrt{1+\sin x}-\sqrt{1-\sin x}\]
\[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)\times \left( \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)\]
Now we know that,
\[\begin{align}
  & \left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}} \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align}\]
So, we have
\[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\dfrac{1+\sin x-1+\sin x}{1+\sin x+1-\sin x-2\sqrt{1-{{\sin }^{2}}x}}\]
\[=\dfrac{2\sin x}{2-2\sqrt{{{\cos }^{2}}x}}\]
\[=\dfrac{2\sin x}{2\left( 1-\cos x \right)}\]
\[=\dfrac{\sin x}{1-\cos x}\]
Now we know the trigonometric identities that
$\begin{align}
  & \sin 2x=2\sin x\cos x \\
 & \Rightarrow \sin x=2\sin \left( \dfrac{x}{2} \right)\cos \left( \dfrac{x}{2} \right)..........(i) \\
 & \cos 2x=1-2{{\sin }^{2}}x \\
 & \Rightarrow \cos x=1-2{{\sin }^{2}}\left( \dfrac{{{x}^{2}}}{2} \right) \\
 & 2{{\sin }^{2}}\left( \dfrac{x}{2} \right)=1-\cos x...........(ii) \\
\end{align}$
Now, we will use equation (i) and (ii) in numerator and denominator
\[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\dfrac{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\sin }^{2}}\dfrac{x}{2}}=\cot \dfrac{x}{2}\]
So, we have
\[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}=\cot \dfrac{x}{2}\]
Now substituting this in y we have,
$y={{\cot }^{-1}}\left( \cot \dfrac{x}{2} \right)$
Also we know that
${{\cot }^{-1}}\left( \cot \theta \right)=\theta $
So,
$y=\dfrac{x}{2}$
Now, we have to find \[\dfrac{dy}{dx}\] . So,
$\dfrac{dy}{dx}=\dfrac{dy}{dx}\left( \dfrac{x}{2} \right)=\dfrac{1}{2}$

So, the correct answer is “Option (a)”.

Note: To solve these types of question one must know trigonometric identities like
\[\begin{align}
  & \sin 2x=2\sin x\cos x \\
 & \cos 2x=2{{\cos }^{2}}x-1 \\
 & {{\cot }^{-1}}\left( \cot x \right)=x \\
\end{align}\]
Also it has to noted the we have rationalized \[\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\] because on doing this the numerator and denominator are both simplified and we get rid of the square root this makes solving the problem easy.