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Question: Prove the following expression \({{\tan }^{-1}}\left( \sqrt{x} \right)=\dfrac{1}{2}{{\cos }^{-1}}\le...

Prove the following expression tan⁡−1(x)=12cos⁡−1(1−x1+x) , x∈[0,1]{{\tan }^{-1}}\left( \sqrt{x} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,,\,x\in \left[ 0,1 \right].

Explanation

Solution

Hint: We will use the substitution x=tan⁡2(θ)x={{\tan }^{2}}\left( \theta \right) here to solve the question further. Using this substitution we will convert the right side of the expression tan⁡−1(x)=12cos⁡−1(1−x1+x) , x∈[0,1]{{\tan }^{-1}}\left( \sqrt{x} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,,\,x\in \left[ 0,1 \right] in terms of inverse tan.

Complete step-by-step answer:
Now we will first consider the expression tan⁡−1(x)=12cos⁡−1(1−x1+x) , x∈[0,1]....(i){{\tan }^{-1}}\left( \sqrt{x} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,,\,x\in \left[ 0,1 \right]....(i).
We will substitute x=tan⁡2(θ)x={{\tan }^{2}}\left( \theta \right) here to the right side of the equation (i). Therefore, we have
12cos⁡−1(1−x1+x) =12cos⁡−1(1−tan⁡2(θ)1+tan⁡2(θ)) \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\left( \theta \right)}{1+{{\tan }^{2}}\left( \theta \right)} \right)\,. As we know that tan⁡(p)=sin⁡(p)cos⁡(p)\tan \left( p \right)=\dfrac{\sin \left( p \right)}{\cos \left( p \right)}. Therefore we will have
12cos⁡−1(1−x1+x) =12cos⁡−1(1−sin⁡2(θ)cos⁡2(θ)1+sin⁡2(θ)cos⁡2(θ))  ⇒12cos⁡−1(1−x1+x) =12cos⁡−1(cos⁡2(θ)−sin⁡2(θ)cos⁡2(θ)cos⁡2(θ)+sin⁡2(θ)cos⁡2(θ))  \begin{aligned} & \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-\dfrac{{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)}}{1+\dfrac{{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)}} \right)\, \\\ & \Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{\dfrac{{{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)}}{\dfrac{{{\cos }^{2}}\left( \theta \right)+{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)}} \right)\, \\\ \end{aligned}
After cancelling the term cos⁡2(θ){{\cos }^{2}}\left( \theta \right) we will have 12cos⁡−1(1−x1+x) =12cos⁡−1(cos⁡2(θ)−sin⁡2(θ)cos⁡2(θ)+sin⁡2(θ))\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{{{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right)}{{{\cos }^{2}}\left( \theta \right)+{{\sin }^{2}}\left( \theta \right)} \right). As we know that cos⁡2(θ)+sin⁡2(θ)=1{{\cos }^{2}}\left( \theta \right)+{{\sin }^{2}}\left( \theta \right)=1. Therefore, we will get
12cos⁡−1(1−x1+x) =12cos⁡−1(cos⁡2(θ)−sin⁡2(θ)1) ⇒12cos⁡−1(1−x1+x) =12cos⁡−1(cos⁡2(θ)−sin⁡2(θ)) \begin{aligned} & \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{{{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right)}{1} \right) \\\ & \Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( {{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right) \right) \\\ \end{aligned}
Now we will apply the formula given by cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)\cos \left( 2\theta \right)={{\cos }^{2}}\left( \theta \right)-{{\sin }^{2}}\left( \theta \right) in the above expression thus, we will get
12cos⁡−1(1−x1+x) =12cos⁡−1(cos⁡(2θ))\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}{{\cos }^{-1}}\left( \cos \left( 2\theta \right) \right)
As we know that cos⁡−1(cos⁡(x))=x{{\cos }^{-1}}\left( \cos \left( x \right) \right)=x therefore, we have
12cos⁡−1(1−x1+x) =12×2θ ⇒12cos⁡−1(1−x1+x) =θ...(ii) \begin{aligned} & \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\dfrac{1}{2}\times 2\theta \\\ & \Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,=\theta ...(ii) \\\ \end{aligned}
Since, we have that x=tan⁡2(θ)x={{\tan }^{2}}\left( \theta \right). By taking root to both the sides of the term x=tan⁡2(θ)x={{\tan }^{2}}\left( \theta \right) we will get x=tan⁡(θ)\sqrt{x}=\tan \left( \theta \right). Now we will take the tan term to the left side of the expression we will get tan⁡−1(x)=θ{{\tan }^{-1}}\left( \sqrt{x} \right)=\theta . Now we will substitute the value of θ\theta in equation (ii). Therefore, we have
12cos⁡−1(1−x1+x) =tan⁡−1(x)\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,={{\tan }^{-1}}\left( \sqrt{x} \right) which is our required expression.
Hence, the expression tan⁡−1(x)=12cos⁡−1(1−x1+x) , x∈[0,1]{{\tan }^{-1}}\left( \sqrt{x} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)\,,\,x\in \left[ 0,1 \right] is proved.

Note: Alternatively we could have used the method in which we can convert cos⁡−1{{\cos }^{-1}} term into tan⁡−1{{\tan }^{-1}}. This can be done by the formula cos⁡−1(x)=tan⁡−1(1−x2x){{\cos }^{-1}}\left( x \right)={{\tan }^{-1}}\left( \dfrac{\sqrt{1-{{x}^{2}}}}{x} \right). We can use this formula here since we have the condition x∈[0,1]x\in \left[ 0,1 \right] which is the required condition for the formula.