Question

Prove the given trigonometric expression
${{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)=\dfrac{x}{2};x\in \left( 0,\dfrac{\pi }{4} \right)$ .

Answer 1

Hint: For solving this question first we will assume $A=\sqrt{1+\sin x}$ and $B=\sqrt{1-\sin x}$. After that, we will use trigonometric formulas like $1={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}$ and $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ for the simplification of $A=\sqrt{1+\sin x}$ and $B=\sqrt{1-\sin x}$ . Then, we will use the formula ${{\cot }^{-1}}\left( \cot x \right)=x\text{ }\left( \text{if }x\in \left( 0,\pi \right) \right)$ for proving the desired result.

Complete step-by-step solution -
Given:
We have to prove the following equation:
${{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)=\dfrac{x}{2};x\in \left( 0,\dfrac{\pi }{4} \right)$
Now, before we proceed we should know the following formulas:
$\begin{aligned} & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.................\left( 1 \right) \\\ & \sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}..............\left( 2 \right) \\\ & {{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}.........\left( 3 \right) \\\ & {{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}.........\left( 4 \right) \\\ & {{\cot }^{-1}}\left( \cot x \right)=x\text{ }\left( \text{if }x\in \left( 0,\pi \right) \right)..............\left( 5 \right) \\\ \end{aligned}$
Now, before we simplify the term ${{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)$ , we should simplify $\sqrt{1+\sin x}$ and $\sqrt{1-\sin x}$ separately.
Now, let $A=\sqrt{1+\sin x}$ and $B=\sqrt{1-\sin x}$ . So, we will use formulas from the above equations to simplify $A$ and $B$ separately.
Simplification of $A=\sqrt{1+\sin x}$ :
Now, we will use the formula from the equation to write $1={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}$ . Then,
$\begin{aligned} & A=\sqrt{1+\sin x} \\\ & \Rightarrow A=\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+\sin x} \\\ \end{aligned}$
Now, we will use the formula from the equation (2) to write $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ in the above equation. Then,
$\begin{aligned} & A=\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+\sin x} \\\ & \Rightarrow A=\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \\\ \end{aligned}$
Now, we will use the formula from the equation (3) to write ${{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}={{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}$ in the above equation. Then,

& A=\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}+2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \\\ & \Rightarrow A=\sqrt{{{\left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right)}^{2}}} \\\ & \Rightarrow A=\left| \left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right) \right| \\\ \end{aligned}$$ Now, as it is given that $x\in \left( 0,\dfrac{\pi }{4} \right)$ so, $\dfrac{x}{2}\in \left( 0,\dfrac{\pi }{8} \right)$ . And we know that, $\sin \theta $ and $\cos \theta $ are positive for $\theta \in \left( 0,\dfrac{\pi }{8} \right)$ . Then, $$\begin{aligned} & A=\left| \left( \sin \dfrac{x}{2}+\cos \dfrac{x}{2} \right) \right| \\\ & \Rightarrow A=\sin \dfrac{x}{2}+\cos \dfrac{x}{2}...................\left( 6 \right) \\\ \end{aligned}$$ Simplification of $B=\sqrt{1-\sin x}$ : Now, we will use the formula from the equation to write $1={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}$ . Then, $\begin{aligned} & B=\sqrt{1-\sin x} \\\ & \Rightarrow B=\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}-\sin x} \\\ \end{aligned}$ Now, we will use the formula from the equation (2) to write $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$ in the above equation. Then, $\begin{aligned} & B=\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}-\sin x} \\\ & \Rightarrow B=\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \\\ \end{aligned}$ Now, we will use the formula from the equation (4) to write ${{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}={{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}^{2}}$ in the above equation. Then, $$\begin{aligned} & B=\sqrt{{{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}-2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \\\ & \Rightarrow B=\sqrt{{{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}^{2}}} \\\ & \Rightarrow B=\left| \left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right) \right| \\\ \end{aligned}$$ Now, as it is given that $x\in \left( 0,\dfrac{\pi }{4} \right)$ so, $\dfrac{x}{2}\in \left( 0,\dfrac{\pi }{8} \right)$ . And we know that $\cos \theta >\sin \theta $ for $\theta \in \left( 0,\dfrac{\pi }{8} \right)$ so, $\cos \theta -\sin \theta $ will be positive for $\theta \in \left( 0,\dfrac{\pi }{8} \right)$ . Then, $$\begin{aligned} & B=\left| \left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right) \right| \\\ & \Rightarrow B=\cos \dfrac{x}{2}-\sin \dfrac{x}{2}...................\left( 7 \right) \\\ \end{aligned}$$ Simplification of ${{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)$ : Now, as per our assumption $A=\sqrt{1+\sin x}$ and $B=\sqrt{1-\sin x}$ . Then, $\begin{aligned} & {{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right) \\\ & \Rightarrow {{\cot }^{-1}}\left( \dfrac{A+B}{A-B} \right) \\\ \end{aligned}$ Now, we will put $$A=\sin \dfrac{x}{2}+\cos \dfrac{x}{2}$$ from equation (6) and $$B=\cos \dfrac{x}{2}-\sin \dfrac{x}{2}$$ from equation (7) in the above expression. Then, $\begin{aligned} & {{\cot }^{-1}}\left( \dfrac{A+B}{A-B} \right) \\\ & \Rightarrow {{\cot }^{-1}}\left( \dfrac{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}+\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}-\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)} \right) \\\ & \Rightarrow {{\cot }^{-1}}\left( \dfrac{2\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}-\cos \dfrac{x}{2}+\sin \dfrac{x}{2}} \right) \\\ & \Rightarrow {{\cot }^{-1}}\left( \dfrac{2\cos \dfrac{x}{2}}{2\sin \dfrac{x}{2}} \right) \\\ & \Rightarrow {{\cot }^{-1}}\left( \dfrac{\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}} \right) \\\ \end{aligned}$ Now, we can write $\dfrac{\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}}=\cot \dfrac{x}{2}$ in the above equation. Then, $\begin{aligned} & {{\cot }^{-1}}\left( \dfrac{\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}} \right) \\\ & \Rightarrow {{\cot }^{-1}}\left( \cot \dfrac{x}{2} \right) \\\ \end{aligned}$ Now, as it is given that $x\in \left( 0,\dfrac{\pi }{4} \right)$ so, $\dfrac{x}{2}\in \left( 0,\dfrac{\pi }{8} \right)$ . So, from formula from equation (5), we can write $${{\cot }^{-1}}\left( \cot \dfrac{x}{2} \right)=\dfrac{x}{2}$$ in the above expression. Then, $$\begin{aligned} & {{\cot }^{-1}}\left( \cot \dfrac{x}{2} \right) \\\ & \Rightarrow \dfrac{x}{2} \\\ \end{aligned}$$ Now, from the above result, we conclude that, ${{\cot }^{-1}}\left( \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right)=\dfrac{x}{2};x\in \left( 0,\dfrac{\pi }{4} \right)$ . Hence, proved. Note: Here, the student should first understand what is asked in the question and then proceed in the right direction to get the correct answer quickly. Moreover, avoid writing $$B=\left| \left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right) \right|=\sin \dfrac{x}{2}-\cos \dfrac{x}{2}$$ , it would be the wrong approach as $$\sin \dfrac{x}{2}-\cos \dfrac{x}{2}$$ will be negative for $x\in \left( 0,\dfrac{\pi }{4} \right)$ . And avoid making calculation mistakes while solving the problem.