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Question: Prove the following result, \({{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cot }^{-1}}\le...

Prove the following result, tan⁡−1(1−x22x)+cot⁡−1(1−x22x)=π2{{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)=\dfrac{\pi }{2}.

Explanation

Solution

Hint:To prove the given equality, we should know few of the inverse trigonometric ratios like, cot⁡−1x=tan⁡−11x{{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x}. Also, we should have some knowledge of the formulas like, tan⁡−1a+tan⁡−1b=tan⁡−1(a+b1−ab){{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right). And we should know that the value of tan⁡π2=∞\tan \dfrac{\pi }{2}=\infty . We can prove the given equality by using these formulas.

Complete step-by-step answer:
In this question, we have been asked to prove that, tan⁡−1(1−x22x)+cot⁡−1(1−x22x)=π2{{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)=\dfrac{\pi }{2}. To prove this, we will first consider the left hand side or the LHS of the given equation. So, we can write it as,
LHS=tan⁡−1(1−x22x)+cot⁡−1(1−x22x)LHS={{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)
Now, we know that cot⁡−1x=tan⁡−11x{{\cot }^{-1}}x={{\tan }^{-1}}\dfrac{1}{x}. So, we can write cot⁡−1(1−x22x){{\cot }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right) as tan⁡−1(2x1−x2){{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right). So, substituting this value in the given equation, we get the LHS as,
LHS=tan⁡−1(1−x22x)+tan⁡−1(2x1−x2)LHS={{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)
Now, we also know that tan⁡−1a+tan⁡−1b=tan⁡−1(a+b1−ab){{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}\left( \dfrac{a+b}{1-ab} \right). So, we will have a=1−x22xa=\dfrac{1-{{x}^{2}}}{2x} and b=2x1−x2b=\dfrac{2x}{1-{{x}^{2}}}. Therefore, applying this in the above equation, we get the LHS as,
LHS=tan⁡−1[(1−x22x)+(2x1−x2)1−(1−x22x)(2x1−x2)]LHS={{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{1-{{x}^{2}}}{2x} \right)+\left( \dfrac{2x}{1-{{x}^{2}}} \right)}{1-\left( \dfrac{1-{{x}^{2}}}{2x} \right)\left( \dfrac{2x}{1-{{x}^{2}}} \right)} \right]
We know that the like terms of the numerator and the denominator get cancelled, so we can write the above equation as follows,
LHS=tan⁡−1[(1−x22x)+(2x1−x2)1−1]LHS={{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{1-{{x}^{2}}}{2x} \right)+\left( \dfrac{2x}{1-{{x}^{2}}} \right)}{1-1} \right]
Which will result in the LHS as,
LHS=tan⁡−1[(1−x22x)+(2x1−x2)0]LHS={{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{1-{{x}^{2}}}{2x} \right)+\left( \dfrac{2x}{1-{{x}^{2}}} \right)}{0} \right]
We know that irrespective of the numerator, if we get 0 in denominator, then the fraction always becomes infinite. So, we can write the LHS as,
LHS=tan⁡−1[∞]LHS={{\tan }^{-1}}\left[ \infty \right]
Now, we know that tan⁡π2=∞\tan \dfrac{\pi }{2}=\infty . So, we can say that, tan⁡−1∞=π2{{\tan }^{-1}}\infty =\dfrac{\pi }{2}. So, we can write the, LHS=π2LHS=\dfrac{\pi }{2}, which is the same as the right hand side or the RHS of the given expression, so LHS = RHS.
Hence, we have proved the given expression.

Note: We should be very careful while solving this question as there are possibilities of calculation mistakes while taking the LCM of terms and simplifying the function. We should also remember that the given expression to be proved is actually an identity tan⁡−1x+cot⁡−1x=π2{{\tan }^{-1}}x+{{\cot }^{-1}}x=\dfrac{\pi }{2}.