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Question: If ab < 1 and \({{\cos }^{-1}}\left( \dfrac{1-{{a}^{2}}}{1+{{a}^{2}}} \right)+{{\cos }^{-1}}\left( \...

If ab < 1 and cos⁡−1(1−a21+a2)+cos⁡−1(1−b21+b2)=2tan⁡−1x{{\cos }^{-1}}\left( \dfrac{1-{{a}^{2}}}{1+{{a}^{2}}} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{b}^{2}}}{1+{{b}^{2}}} \right)=2{{\tan }^{-1}}x , then the value of ‘x’ is equal to?
A.a1+ab\dfrac{a}{1+ab}
B.a1−ab\dfrac{a}{1-ab}
C.a−b1+ab\dfrac{a-b}{1+ab}
D.a+b1+ab\dfrac{a+b}{1+ab}
E.a+b1−ab\dfrac{a+b}{1-ab}

Explanation

Solution

Hint: Use substitutions to convert the equation in terms of ‘tan’ and then use the formula of half angle i.e. 1−tan⁡2x1+tan⁡2x=cos⁡2x\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\cos 2x. Then use the formula cos⁡−1(cos⁡x)=x{{\cos }^{-1}}\left( \cos x \right)=x and replace the substituted values by original values. Then use the formula tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy} to get the final answer.

Complete step by step answer:
To find the value of ‘x’ we have to write down the equation given in the problem, therefore,
cos⁡−1(1−a21+a2)+cos⁡−1(1−b21+b2)=2tan⁡−1x{{\cos }^{-1}}\left( \dfrac{1-{{a}^{2}}}{1+{{a}^{2}}} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{b}^{2}}}{1+{{b}^{2}}} \right)=2{{\tan }^{-1}}x
To simplify the above equation we have to use substitutions therefore,
Put,
a=tan⁡θa=\tan \theta And b=tan⁡βb=\tan \beta …………………………………………………… (1)
Therefore,
θ=tan⁡−1a\theta ={{\tan }^{-1}}a And β=tan⁡−1b\beta ={{\tan }^{-1}}b …………………………………………………. (2)
If we substitute the substitutions of equation (1) in the given equation we will get,
cos⁡−1(1−tan⁡2θ1+tan⁡2θ)+cos⁡−1(1−tan⁡2β1+tan⁡2β)=2tan⁡−1x{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)+{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\beta }{1+{{\tan }^{2}}\beta } \right)=2{{\tan }^{-1}}x
To proceed further in the solution we should know the formula given below,
Formula:
1−tan⁡2x1+tan⁡2x=cos⁡2x\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\cos 2x
By using the formula given above we can write the given equation as follows,
cos⁡−1(cos⁡2θ)+cos⁡−1(cos⁡2β)=2tan⁡−1x{{\cos }^{-1}}\left( \cos 2\theta \right)+{{\cos }^{-1}}\left( \cos 2\beta \right)=2{{\tan }^{-1}}x
Now, to simplify the equation we should know the formula given below,
Formula:
cos⁡−1(cos⁡x)=x{{\cos }^{-1}}\left( \cos x \right)=x
By using the above formula we will get,
2θ+2β=2tan⁡−1x2\theta +2\beta =2{{\tan }^{-1}}x
Taking 2 common from the above equation we will get,
2(θ+β)=2tan⁡−1x2\left( \theta +\beta \right)=2{{\tan }^{-1}}x
We can easily cancel out 2 from both sides therefore we will get,
θ+β=tan⁡−1x\theta +\beta ={{\tan }^{-1}}x
Now we will put the value of equation (2) in the above equation, therefore we will get,
tan⁡−1a+tan⁡−1b=tan⁡−1x{{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}x ………………………………………… (3)
To proceed further in the solution we should know the formula given below,
Formula:
tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy} , where xy < 1
As we have given in the problem that ab < 1 therefore we can use the above formula.
Therefore by using the above formula in the equation (3) we will get,
tan⁡−1a+b1−ab=tan⁡−1x{{\tan }^{-1}}\dfrac{a+b}{1-ab}={{\tan }^{-1}}x
As there is tan⁡−1{{\tan }^{-1}} on both sides of the equation and therefore we can cancel it out from the equation, therefore we will get,
∴a+b1−ab=x\therefore \dfrac{a+b}{1-ab}=x
By rearranging the above equation we will get,
∴x=a+b1−ab\therefore x=\dfrac{a+b}{1-ab}
Therefore the value of ‘x’ is equal to a+b1−ab\dfrac{a+b}{1-ab}.
Therefore the correct answer is option (e).

Note: While using the formula tan⁡−1x+tan⁡−1y=tan⁡−1x+y1−xy{{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy} do remember that it will obey only if xy < 1 and therefore do check the given conditions so that you can avoid a silly mistake.