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Question: If x and y are positive and xy > 1, then what is \({{\tan }^{-1}}x+{{\tan }^{-1}}y\) equal to? a.\...

If x and y are positive and xy > 1, then what is tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y equal to?
a.tan⁡−1(x+y1−xy){{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)
b.π+tan⁡−1(x+y1−xy)\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)
c.π−tan⁡−1(x+y1−xy)\pi -{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)
d.tan⁡−1(x−y1+xy){{\tan }^{-1}}\left( \dfrac{x-y}{1+xy} \right)

Explanation

Solution

Hint: First assume the given terms as variables and write the values of ‘x’ and ‘y’ in terms of those variables. Then use the formula tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} and put the values of the variables and ‘x’ and ‘y’ to get the value of given expression. Then check whether the answer is true for xy > 1 if not then add “π\pi ” in it as ‘tan’ is positive in the third quadrant to get the final answer.

Complete step-by-step answer:

To solve the given question we will write the given expression first, therefore,
∴tan⁡−1x+tan⁡−1y\therefore {{\tan }^{-1}}x+{{\tan }^{-1}}y
To simplify the above equation we will assume,
tan⁡−1x=a{{\tan }^{-1}}x=a and tan⁡−1y=b{{\tan }^{-1}}y=b ………………………………………….. (1)
If we shift ‘tan⁡−1{{\tan }^{-1}}’ on the right hand side we will get,
x=tan⁡ax=\tan a and y=tan⁡by=\tan b ……………………………………………….. (2)
To proceed further in the solution we should know the formula given below,
Formula:
tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}
By using the above formula we can write the formula for ‘tan (a + b) as follows,
∴tan⁡(a+b)=tan⁡a+tan⁡b1−tan⁡atan⁡b\therefore \tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}
Now if we put the values of equation (2) in the above equation we will get,
∴tan⁡(a+b)=x+y1−xy\therefore \tan \left( a+b \right)=\dfrac{x+y}{1-xy}
If we shift ‘tan’ on the right hand side of the above equation we will get,
∴a+b=tan⁡−1(x+y1−xy)\therefore a+b={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)
If we put the values of equation (2) in the above equation we will get,
∴tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy)\therefore {{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) ……………………………………………… (3)
Now, as we have given the condition that, x is positive, y is positive and xy > 1 therefore we can write it as,
x > 0, y > 0 and xy > 1
Consider the left hand side of the equation (3),
Left Hand Side (LHS) = tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y
As x > 0 and y > 0 therefore the value of above equation will become positive therefore we will get a positive angle,
Left Hand Side (LHS) = (+ ve) + (+ ve) = + ve value …………………………………. (4)
Now consider the right hand side of the equation (3),
Right Hand Side = tan⁡−1(x+y1−xy){{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)
As x > 0 and y > 0 therefore the value of x + y will also become positive, but as xy > 1 therefore the value of 1 – xy will become negative, therefore above equation will become,
Right Hand Side = tan⁡−1(+ve−ve){{\tan }^{-1}}\left( \dfrac{+ve}{-ve} \right)
Therefore, Right Hand Side = tan⁡−1(−ve){{\tan }^{-1}}\left( -ve \right)
As we know tan⁡−1(−x)=−tan⁡−1x{{\tan }^{-1}}\left( -x \right)=-{{\tan }^{-1}}x therefore the above value will become negative,
Therefore, Right Hand Side = - ve value.
If we compare above equation with equation (4) then we can say that both are not equal and therefore the identity is false when xy > 1,
∴tan⁡−1x+tan⁡−1y≠tan⁡−1(x+y1−xy)\therefore {{\tan }^{-1}}x+{{\tan }^{-1}}y\ne {{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) when , x > 0, y > 0 and xy > 1
Now as we know that ‘tan’ is positive in the first and third quadrant and as the value of the above equation in the first quadrant is becoming negative therefore it should lie in the third quadrant.
Therefore if we add the right hand side of the equation (4) by “π\pi ” then we will get the positive answer as it will enter in the third quadrant. Therefore equation (4) will become,
∴tan⁡−1x+tan⁡−1y=π+tan⁡−1(x+y1−xy)\therefore {{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) when x > 0, y > 0, and xy > 1
Therefore if x and y are positive and xy > 1 then the value of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y is equal to π+tan⁡−1(x+y1−xy)\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right).
Therefore the correct answer is option (b).

Note: The step tan⁡(a+b)=tan⁡a+tan⁡b1−tan⁡atan⁡b\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b} is the key point to complete solution therefore try to remember it. Also many students won’t think of the condition given i.e. xy > 1 and they directly write the answer as tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right) which is a silly mistake.