Solveeit Logo
Login

Question

Question: Prove the following statement, \(\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{...

Prove the following statement, sin⁡[tan⁡−1(1−x22x)+cos⁡−1(1−x21+x2)]=1\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right]=1.

Explanation

Solution

Hint:In order to solve this question, we should have some knowledge of the inverse trigonometric formulas like, cos⁡−1(1−x21+x2)=2tan⁡−1x{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x and 2tan⁡−1x=tan⁡−1(2x1−x2)2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) and tan⁡−1x=cot⁡−11x{{\tan }^{-1}}x={{\cot }^{-1}}\dfrac{1}{x} and cot⁡−1x+tan⁡−1x=π2{{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}. By using these formulas, we can prove the desired relation.

Complete step-by-step answer:
In the given question, we have been asked to prove that, sin⁡[tan⁡−1(1−x22x)+cos⁡−1(1−x21+x2)]=1\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right]=1. To prove the same, we will first consider the left hand side or the LHS of the equality. So, we can write the LHS as,
LHS=sin⁡[tan⁡−1(1−x22x)+cos⁡−1(1−x21+x2)]LHS=\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right) \right]
Now, we know that, cos⁡−1(1−x21+x2)=2tan⁡−1x{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)=2{{\tan }^{-1}}x. So, we can apply it here in the above equality and get the LHS as follows,
LHS=sin⁡[tan⁡−1(1−x22x)+2tan⁡−1x]LHS=\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+2{{\tan }^{-1}}x \right]
Now, we know that 2tan⁡−1x2{{\tan }^{-1}}x can be expressed as tan⁡−1(2x1−x2){{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right). So, we will substitute it in the above equality. So, we get the LHS as,
LHS=sin⁡[tan⁡−1(1−x22x)+tan⁡−1(2x1−x2)]LHS=\sin \left[ {{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) \right]
Now, we also know that tan⁡−1x=cot⁡−11x{{\tan }^{-1}}x={{\cot }^{-1}}\dfrac{1}{x}. So, we can write tan⁡−1(1−x22x){{\tan }^{-1}}\left( \dfrac{1-{{x}^{2}}}{2x} \right) as cot⁡−1(2x1−x2){{\cot }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right). Therefore, we can substitute it in the LHS. So, we get,
LHS=sin⁡[cot⁡−1(2x1−x2)+tan⁡−1(2x1−x2)]LHS=\sin \left[ {{\cot }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)+{{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) \right]
Now, we know that cot⁡−1x+tan⁡−1x=π2{{\cot }^{-1}}x+{{\tan }^{-1}}x=\dfrac{\pi }{2}. So, by using this identity, we can write the LHS as,
LHS=sin⁡(π2)LHS=\sin \left( \dfrac{\pi }{2} \right)
We know that the sine ratio of π2\dfrac{\pi }{2} is 1, that is sin⁡(π2)=1\sin \left( \dfrac{\pi }{2} \right)=1. So, we can write the LHS as,
LHS=1LHS=1.
We can see that it is the same as the right hand side or the RHS of the given equality. Therefore LHS = RHS.
Hence, we have proved the given equality.

Note: While solving the given question, one can think of applying the identity of tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) in the step, where we applied tan⁡−1x=cot⁡−11x{{\tan }^{-1}}x={{\cot }^{-1}}\dfrac{1}{x}. This would not be incorrect, but it is a lengthy method, so it is better to avoid this and leave the solution simpler.