Question

If ab < 1 and ${{\cos }^{-1}}\left( \dfrac{1-{{a}^{2}}}{1+{{a}^{2}}} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{b}^{2}}}{1+{{b}^{2}}} \right)=2{{\tan }^{-1}}x$ , then the value of ‘x’ is equal to?
A.$\dfrac{a}{1+ab}$
B.$\dfrac{a}{1-ab}$
C.$\dfrac{a-b}{1+ab}$
D.$\dfrac{a+b}{1+ab}$
E.\[\dfrac{a+b}{1-ab}\]

Answer 1

Hint: Use substitutions to convert the equation in terms of ‘tan’ and then use the formula of half angle i.e. $\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\cos 2x$. Then use the formula ${{\cos }^{-1}}\left( \cos x \right)=x$ and replace the substituted values by original values. Then use the formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy}$ 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,
${{\cos }^{-1}}\left( \dfrac{1-{{a}^{2}}}{1+{{a}^{2}}} \right)+{{\cos }^{-1}}\left( \dfrac{1-{{b}^{2}}}{1+{{b}^{2}}} \right)=2{{\tan }^{-1}}x$
To simplify the above equation we have to use substitutions therefore,
Put,
$a=\tan \theta $ And $b=\tan \beta $ …………………………………………………… (1)
Therefore,
$\theta ={{\tan }^{-1}}a$ And $\beta ={{\tan }^{-1}}b$ …………………………………………………. (2)
If we substitute the substitutions of equation (1) in the given equation we will get,
 ${{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right)+{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\beta }{1+{{\tan }^{2}}\beta } \right)=2{{\tan }^{-1}}x$
To proceed further in the solution we should know the formula given below,
Formula:
$\dfrac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}=\cos 2x$
By using the formula given above we can write the given equation as follows,
${{\cos }^{-1}}\left( \cos 2\theta \right)+{{\cos }^{-1}}\left( \cos 2\beta \right)=2{{\tan }^{-1}}x$
Now, to simplify the equation we should know the formula given below,
Formula:
${{\cos }^{-1}}\left( \cos x \right)=x$
By using the above formula we will get,
$2\theta +2\beta =2{{\tan }^{-1}}x$
Taking 2 common from the above equation we will get,
$2\left( \theta +\beta \right)=2{{\tan }^{-1}}x$
We can easily cancel out 2 from both sides therefore we will get,
$\theta +\beta ={{\tan }^{-1}}x$
Now we will put the value of equation (2) in the above equation, therefore we will get,
${{\tan }^{-1}}a+{{\tan }^{-1}}b={{\tan }^{-1}}x$ ………………………………………… (3)
To proceed further in the solution we should know the formula given below,
Formula:
${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy}$ , 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,
${{\tan }^{-1}}\dfrac{a+b}{1-ab}={{\tan }^{-1}}x$
As there is ${{\tan }^{-1}}$ on both sides of the equation and therefore we can cancel it out from the equation, therefore we will get,
$\therefore \dfrac{a+b}{1-ab}=x$
By rearranging the above equation we will get,
$\therefore x=\dfrac{a+b}{1-ab}$
Therefore the value of ‘x’ is equal to $\dfrac{a+b}{1-ab}$.
Therefore the correct answer is option (e).

Note: While using the formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\dfrac{x+y}{1-xy}$ 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.