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Question: Prove that \[{{\tan }^{-1}}\sqrt{x}=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right),x\in \...

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

Explanation

Solution

Hint: In this question, we first need to assume some value for the left-hand side of the given function. Then get the value of x in terms of variable assumed. Now, find the value of 1−x1-x and 1+x1+x then divide them and simplify further to get the relation.

Complete step-by-step solution -
From the given equation in the question we have
tan⁡−1x=12cos⁡−1(1−x1+x){{\tan }^{-1}}\sqrt{x}=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)
Now, from the left hand side of the given equation in the question we have
⇒tan⁡−1x\Rightarrow {{\tan }^{-1}}\sqrt{x}
Now, let us assume this expression as some theta
⇒tan⁡−1x=θ\Rightarrow {{\tan }^{-1}}\sqrt{x}=\theta
Now, let us apply the tangent function on both the sides to simplify it further
⇒x=tan⁡θ\Rightarrow \sqrt{x}=\tan \theta
Let us now do the squaring on both sides
⇒(x)2=(tan⁡θ)2\Rightarrow {{\left( \sqrt{x} \right)}^{2}}={{\left( \tan \theta \right)}^{2}}
Now, on simplifying it further we get,
⇒x=tan⁡2θ\Rightarrow x={{\tan }^{2}}\theta
Let us now find the values of the expressions given in the left hand side
Let us first find the value of 1−x1-x
⇒1−x\Rightarrow 1-x
Now, on substituting the value of x in the above expression we get,
⇒1−x=1−tan⁡2θ\Rightarrow 1-x=1-{{\tan }^{2}}\theta
Let us now find the value of 1+x1+x
⇒1+x\Rightarrow 1+x
Now, on substituting the value of x we get,
⇒1+x=1+tan⁡2θ\Rightarrow 1+x=1+{{\tan }^{2}}\theta
Let us now divide these two equations and simplify further
⇒1−x1+x=1−tan⁡2θ1+tan⁡2θ\Rightarrow \dfrac{1-x}{1+x}=\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }
As we already know from the trigonometric identities that
1+tan⁡2θ=sec⁡2θ1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta
Now, on using the above identity and simplifying further we get,
⇒1−x1+x=1−tan⁡2θsec⁡2θ\Rightarrow \dfrac{1-x}{1+x}=\dfrac{1-{{\tan }^{2}}\theta }{{{\sec }^{2}}\theta }
Now, this can also be written as
⇒1−x1+x=1sec⁡2θ−tan⁡2θsec⁡2θ\Rightarrow \dfrac{1-x}{1+x}=\dfrac{1}{{{\sec }^{2}}\theta }-\dfrac{{{\tan }^{2}}\theta }{{{\sec }^{2}}\theta }
Now, using the trigonometric identities and simplifying it further we get,
⇒1−x1+x=cos⁡2θ−sin⁡2θ\Rightarrow \dfrac{1-x}{1+x}={{\cos }^{2}}\theta -{{\sin }^{2}}\theta
Now, this can also be written as
⇒1−x1+x=cos⁡2θ\Rightarrow \dfrac{1-x}{1+x}=\cos 2\theta
Now, let us apply inverse of cosine function on both the sides
⇒cos⁡−1(1−x1+x)=2θ\Rightarrow {{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)=2\theta
Now, on further rearranging the terms we get,
∴θ=12cos⁡−1(1−x1+x)\therefore \theta =\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right)
Hence,
tan⁡−1x=12cos⁡−1(1−x1+x){{\tan }^{-1}}\sqrt{x}=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-x}{1+x} \right).

Note: Instead of assuming the function to be variable and then solving we can also solve it by using the properties of inverse trigonometric functions by first converting the given inverse tangent function into an inverse cosine function and simplify it further accordingly. It is important to note that while solving the given equations by substituting the respective values we should not neglect any of the terms or interchange the numerator and denominator because it changes the result completely.