Question

Find the value of ${{\cot }^{-1}}\left( \dfrac{\sqrt{1-\sin x}-\sqrt{1+\sin x}}{\sqrt{1-\sin x}+\sqrt{1+\sin x}} \right)=...\left( 0 < x < \dfrac{\pi }{2} \right)$.
$\begin{aligned} & \left. A \right)\dfrac{x}{2} \\\ & \left. B \right)\dfrac{\pi }{2}-2x \\\ & \left. C \right)2\pi -x \\\ & \left. D \right)\pi -\dfrac{x}{2} \\\ \end{aligned}$

Answer 1

In this question, we have to find the value of given trigonometric function. Thus, we will use the trigonometric formulas and the identities to get the solution. First, we will use the trigonometric identity ${{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}=1$ and trigonometric formula $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ in the given question. After that, we will use the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ such that $a=\sin \dfrac{x}{2}$ and $b=\cos \dfrac{x}{2}$ . Then, we will divide $\sin \dfrac{x}{2}$ in both the numerator and the denominator. In the last, we will apply the cot function formula, that is $\cot \left( a+b \right)=\dfrac{\cot \left( a \right)\cot \left( b \right)-1}{\cot a+\cot b}$ . In the last, we will use the formula ${{\cot }^{-1}}\left( \cot x \right)=x$, to get the required solution.

Complete step by step solution:
According to the problem, we have to find the value of given trigonometric function.
Thus, we will use the trigonometric formulas and the identities to get the solution.
The trigonometric function given to us is ${{\cot }^{-1}}\left( \dfrac{\sqrt{1-\sin x}-\sqrt{1+\sin x}}{\sqrt{1-\sin x}+\sqrt{1+\sin x}} \right)$ --------- (1)
Now, we will first use the trigonometric identity ${{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}=1$ and trigonometric formula $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ in equation (1), we get
$\Rightarrow {{\cot }^{-1}}\left( \dfrac{\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}-\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}+\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}} \right)$
Now, we will apply the algebraic identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ and ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ such that $a=\sin \dfrac{x}{2}$ and $b=\cos \dfrac{x}{2}$ in the above equation, we get
$\Rightarrow {{\cot }^{-1}}\left( \dfrac{\sqrt{{{\left( \sin \dfrac{x}{2}-\cos \dfrac{x}{2} \right)}^{2}}}-\sqrt{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}}{\sqrt{{{\left( \sin \dfrac{x}{2}-\cos \dfrac{x}{2} \right)}^{2}}}+\sqrt{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}}} \right)$
Now, we will solve the above expression more, we get
$\Rightarrow {{\cot }^{-1}}\left( \dfrac{\sin \dfrac{x}{2}-\cos \dfrac{x}{2}-\sin \dfrac{x}{2}-\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}-\cos \dfrac{x}{2}+\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)} \right)$
As we know, the same terms with opposite signs cancel out each, we get
$\Rightarrow {{\cot }^{-1}}\left( \dfrac{-\cos \dfrac{x}{2}-\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}+\sin \dfrac{x}{2}} \right)$
$\Rightarrow {{\cot }^{-1}}\left( \dfrac{-2\cos \dfrac{x}{2}}{2\sin \dfrac{x}{2}} \right)$
$\Rightarrow {{\cot }^{-1}}\left( \dfrac{-\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}} \right)$
Now, we will apply the trigonometric formula $\cot x=\dfrac{\cos x}{\sin x}$ in the above expression, we get
$\Rightarrow {{\cot }^{-1}}\left( -\cot \dfrac{x}{2} \right)$
Now, we know that cot is an odd function, thus we will apply the formula $-\cot x=\cot \left( -x \right)$ in the above expression, we get
$\Rightarrow {{\cot }^{-1}}\left( \cot \left( -\dfrac{x}{2} \right) \right)$
Now, we know that x must lie between 0 and $\dfrac{\pi }{2}$ , therefore it is the first quadrant where all the functions are positive, hence we rewrite $\cot \left( -x \right)=\cot \left( \pi -x \right)$ , we get
$\Rightarrow {{\cot }^{-1}}\left( \cot \left( \pi -\dfrac{x}{2} \right) \right)$
Now, we will apply the inverse-trigonometric formula ${{\cot }^{-1}}\left( \cot x \right)=x$ in the above expression, we get
$\Rightarrow \pi -\dfrac{x}{2}$
Therefore, for ${{\cot }^{-1}}\left( \dfrac{\sqrt{1-\sin x}-\sqrt{1+\sin x}}{\sqrt{1-\sin x}+\sqrt{1+\sin x}} \right)=...\left( 0 < x < \dfrac{\pi }{2} \right)$ , it is equal to $\pi -\dfrac{x}{2}$

So, the correct answer is “Option D”.

Note: While solving this problem, do mention all the steps properly to avoid error and confusion. Do mention all the formulas properly and do not forget to change the negative angle to positive angle, to get an accurate solution.