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Question: Prove that \({{\cot }^{-1}}\left[ \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\s...

Prove that cot⁡−1[1+sin⁡x−1−sin⁡x1+sin⁡x+1−sin⁡x]=[π2−x2];x∈(0,π4){{\cot }^{-1}}\left[ \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right]=\left[\dfrac{\pi}{2}-\dfrac{x}{2} \right];x\in \left( 0,\dfrac{\pi }{4} \right).

Explanation

Solution

Change the given cot inverse function into tan inverse function by using the formula: cot⁡−1x=tan⁡−11x{{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x}. Now, write sin⁡x=2sin⁡x2cos⁡x2\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} and 1=cos⁡2x2+sin⁡2x21={{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}. Using these substitutions, write, 1+sin⁡x=(cos⁡x2+sin⁡x2)21+\sin x={{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}} and 1−sin⁡x=(cos⁡x2−sin⁡x2)21-\sin x={{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}^{2}}. Remove the square root sign. A modulus sign will be introduced. To remove modulus signs, check whether the expression inside the mod is positive or negative. If it is positive then, remove the mod simply and if it is negative then remove the mod by adding a negative sign in the expression. Finally, use the identity: tan⁡−1(tan⁡θ)=θ{{\tan }^{-1}}\left( \tan \theta \right)=\theta , to get the answer.

Complete step-by-step solution:
We have to prove: cot⁡−1[1+sin⁡x−1−sin⁡x1+sin⁡x+1−sin⁡x]=x2{{\cot }^{-1}}\left[ \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right]=\dfrac{x}{2}
L.H.S=cot⁡−1[1+sin⁡x−1−sin⁡x1+sin⁡x+1−sin⁡x]L.H.S={{\cot }^{-1}}\left[ \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right]
Converting the given cot inverse function into tan inverse function by using the formula: cot⁡−1x=tan⁡−11x{{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x}, we get,
L.H.S=cot⁡−1[1+sin⁡x−1−sin⁡x1+sin⁡x+1−sin⁡x]=tan⁡−1[1+sin⁡x+1−sin⁡x1+sin⁡x−1−sin⁡x]L.H.S={{\cot }^{-1}}\left[ \dfrac{\sqrt{1+\sin x}-\sqrt{1-\sin x}}{\sqrt{1+\sin x}+\sqrt{1-\sin x}} \right]={{\tan }^{-1}}\left[ \dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}} \right]
Now, writing sin⁡x=2sin⁡x2cos⁡x2\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2} and 1=cos⁡2x2+sin⁡2x21={{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}, we get,
L.H.S=tan⁡−1[sin⁡2x2+cos⁡2x2+2sin⁡x2cos⁡x2+sin⁡2x2+cos⁡2x2−2sin⁡x2cos⁡x2sin⁡2x2+cos⁡2x2+2sin⁡x2cos⁡x2−sin⁡2x2+cos⁡2x2−2sin⁡x2cos⁡x2]L.H.S={{\tan }^{-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]
This can be written as,
L.H.S=tan⁡−1[(cos⁡x2+sin⁡x2)2+(cos⁡x2−sin⁡x2)2(cos⁡x2+sin⁡x2)2−(cos⁡x2−sin⁡x2)2] =tan⁡−1[∣cos⁡x2+sin⁡x2∣+∣cos⁡x2−sin⁡x2∣∣cos⁡x2+sin⁡x2∣−∣cos⁡x2−sin⁡x2∣] \begin{aligned} & L.H.S={{\tan }^{-1}}\left[ \dfrac{\sqrt{{{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}}+\sqrt{{{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}^{2}}}}{\sqrt{{{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}}-\sqrt{{{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}^{2}}}} \right] \\\ & ={{\tan }^{-1}}\left[ \dfrac{\left| \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right|+\left| \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right|}{\left| \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right|-\left| \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right|} \right] \\\ \end{aligned}
Now, we have been given x∈(0,π4)x\in \left( 0,\dfrac{\pi }{4} \right). Therefore, x2∈(0,π8)\dfrac{x}{2}\in \left( 0,\dfrac{\pi }{8} \right).
We know that, in this range cosine of any angle is greater than sine of that angle. Therefore, the terms inside mod are positive.
⇒L.H.S=tan⁡−1[(cos⁡x2+sin⁡x2)+(cos⁡x2−sin⁡x2)(cos⁡x2+sin⁡x2)−(cos⁡x2−sin⁡x2)] =tan⁡−1[2cos⁡x22sin⁡x2] =tan⁡−1[cos⁡x2sin⁡x2] =tan⁡−1[cot⁡x2] =tan⁡−1[tan⁡(π2−x2)] \begin{aligned} & \Rightarrow L.H.S={{\tan }^{-1}}\left[ \dfrac{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)+\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)}{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)-\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)} \right] \\\ & ={{\tan }^{-1}}\left[ \dfrac{2\cos \dfrac{x}{2}}{2\sin \dfrac{x}{2}} \right] \\\ & ={{\tan }^{-1}}\left[ \dfrac{\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}} \right] \\\ & ={{\tan }^{-1}}\left[ \cot \dfrac{x}{2} \right] \\\ & ={{\tan }^{-1}}\left[ \tan\left(\dfrac{\pi}{2}- \dfrac{x}{2}\right) \right] \\\ \end{aligned}
Using the identity: tan⁡−1(tan⁡θ)=θ{{\tan }^{-1}}\left( \tan \theta \right)=\theta , we get,
L.H.S=[π2−x2] =R.H.S \begin{aligned} & L.H.S=\left[\dfrac{\pi}{2}-\dfrac{x}{2} \right] \\\ & =R.H.S \\\ \end{aligned}

Note: One may note that while removing the mod, we have to be careful about the conditions given, that is, the range of ‘x’. Positive or negative value of a trigonometric function depends on the quadrant in which the angle is lying. In the above question, the angle x2\dfrac{x}{2} lies in the first quadrant, therefore, the value of the trigonometric function inside the mod is positive.