Question

If a function is given as $y=\dfrac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}$, then $\dfrac{dy}{dx}$ is equal to
 (a) $-\dfrac{1}{2}{{sec}^{2}}\dfrac{x}{2}$
(b) $-\dfrac{1}{2}{{sec}}\dfrac{x}{2}$
(c) $-\dfrac{1}{2}{{sec}^{2}}{x}$
(d) $-{{sec}^{2}}\dfrac{x}{2}$

Answer 1

First solve $\left( 1-\sin x \right)$ and $\left( 1+\sin x \right)$, then substitute in the given expression and simplify the expression. Then apply the Quotient rule of differentiation.

Complete step-by-step solution -
Given,
$y=\dfrac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}.........(i)$
First we will solve $\left( 1-\sin x \right)$ and $\left( 1+\sin x \right)$ separately.
We know, ${{\sin }^{2}}x+{{\cos }^{2}}x=1,\sin 2x=2\sin x\cos x$
So,
\[1-\sin x={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}\]
And ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$, so
\[1-\sin x={{\left( \sin \dfrac{x}{2}-\cos \dfrac{x}{2} \right)}^{2}}\ldots \ldots .\left( ii \right)\]
Similarly,
\[1+\sin x={{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}\ldots \ldots .\left( iii \right)\]
Substituting equation (ii) and (iii) in equation (i), we get
\[y=\dfrac{\sqrt{{{\left( \sin \dfrac{x}{2}-\cos \dfrac{x}{2} \right)}^{2}}}+\sqrt{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}}{\sqrt{{{\left( \sin \dfrac{x}{2}-\cos \dfrac{x}{2} \right)}^{2}}}-\sqrt{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}}\]
\[\Rightarrow y=\dfrac{\sin \dfrac{x}{2}-\cos \dfrac{x}{2}+\sin \dfrac{x}{2}+\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}-\cos \dfrac{x}{2}-\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}\]
\[\Rightarrow y=\dfrac{2\sin \dfrac{x}{2}}{-2\cos \dfrac{x}{2}}\]
\[\Rightarrow y=-\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}\]
Now we will differentiate with respect to x, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( -\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} \right)\]
Now we will apply the quotient rule, i.e., $\left( \dfrac{u\left( x \right)}{v\left( x \right)} \right)'=\dfrac{{u}'\left( x \right).v\left( x \right)-u\left( x \right).{v}'\left( x \right)}{v{{\left( x \right)}^{2}}}$, so
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{\dfrac{d}{dx}\left( \sin \dfrac{x}{2} \right)\times \left( \cos \dfrac{x}{2} \right)-\left( \sin \dfrac{x}{2} \right)\dfrac{d}{dx}\left( \cos \dfrac{x}{2} \right)}{{{\left( \cos \dfrac{x}{2} \right)}^{2}}}\]
 We know differentiation of $\sin x$ and $\cos x$ is $\cos x$ and $-\sin x$, so above equation becomes,
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{\cos \dfrac{x}{2}\dfrac{d}{dx}\left( \dfrac{x}{2} \right)\times \left( \cos \dfrac{x}{2} \right)-\left( \sin \dfrac{x}{2} \right)\left( -\sin \dfrac{x}{2} \right)\dfrac{d}{dx}\left( \dfrac{x}{2} \right)}{{{\left( \cos \dfrac{x}{2} \right)}^{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{{{\cos }^{2}}\dfrac{x}{2}\left( \dfrac{1}{2} \right)+\left( {{\sin }^{2}}\dfrac{x}{2} \right)\left( \dfrac{1}{2} \right)}{{{\left( \cos \dfrac{x}{2} \right)}^{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{\dfrac{1}{2}\left( {{\cos }^{2}}\dfrac{x}{2}+\left( {{\sin }^{2}}\dfrac{x}{2} \right) \right)}{{{\left( \cos \dfrac{x}{2} \right)}^{2}}}\]
But $\left( {{\sin }^{2}}x+{{\cos }^{2}}x=1 \right)$, so above equation becomes,
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{\dfrac{1}{2}\left( 1 \right)}{{{\left( \cos \dfrac{x}{2} \right)}^{2}}}\]
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{2{{\cos }^{2}}\dfrac{x}{2}}\]
But we know, $\sec x=\dfrac{1}{\cos x}$, so above equation becomes,
\[\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{2}{{\sec }^{2}}x\]
Hence, the correct option for the given question is option (a).
So,the answer is option (a).

Note: In this seeing the question we will first rationalize the given equation and try to find the differentiation. It is possible to get the answer but the solution process will be lengthy.