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Question: If \(x + y = 1,xy = 2\) then \({\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \) A) \(\dfrac{\pi }{4}\) B) ...

If x+y=1,xy=2x + y = 1,xy = 2 then tan⁡−1x+tan⁡−1y={\tan ^{ - 1}}x + {\tan ^{ - 1}}y =
A) π4\dfrac{\pi }{4}
B) 3π4\dfrac{{3\pi }}{4}
C) 3π2\dfrac{{3\pi }}{2}
D) π2\dfrac{\pi }{2}

Explanation

Solution

Apply the inverse formula tan⁡−1A+tan⁡−1B=nπ+tan⁡−1A+B1−AB{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = n\pi + {\tan ^{ - 1}}\dfrac{{A + B}}{{1 - AB}}, where n is a suitably chosen integer. After that substitute the values in the formula and simplify the term. Then replace the value in terms of tan. After that cancel out tan⁡−1{\tan ^{ - 1}} and tan⁡\tan to get the desired result.

Complete step-by-step answer:
The inverse trigonometric functions perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. We know that trig functions are especially applicable to the right-angle triangle. These six important functions are used to find the angle measure in the right triangle when two sides of the triangle measures are known.
The convention symbol to represent the inverse trigonometric function using arc-prefix like arcsin(x), arccos(x), arctan(x), arccsc(x), arcsec(x), arccot(x).
We know that,
tan⁡−1A+tan⁡−1B=nπ+tan⁡−1A+B1−AB{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = n\pi + {\tan ^{ - 1}}\dfrac{{A + B}}{{1 - AB}}
Substitute A=xA = x and B=yB = y in the formula,
⇒tan⁡−1x+tan⁡−1y=nπ+tan⁡−1x+y1−xy\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = n\pi + {\tan ^{ - 1}}\dfrac{{x + y}}{{1 - xy}}
Substitute the values,
⇒tan⁡−1x+tan⁡−1y=nπ+tan⁡−111−2\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = n\pi + {\tan ^{ - 1}}\dfrac{1}{{1 - 2}}
Subtract the terms in the denominator,
⇒tan⁡−1x+tan⁡−1y=nπ+tan⁡−11−1\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = n\pi + {\tan ^{ - 1}}\dfrac{1}{{ - 1}}
Simplify the terms,
⇒tan⁡−1x+tan⁡−1y=nπ+tan⁡−1−1\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = n\pi + {\tan ^{ - 1}} - 1
We know that,
tan⁡−1(−x)=−tan⁡−1x{\tan ^{ - 1}}\left( { - x} \right) = - {\tan ^{ - 1}}x
Use the identity in the above equation,
⇒tan⁡−1x+tan⁡−1y=nπ−tan⁡−11\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = n\pi - {\tan ^{ - 1}}1
Substitute tan⁡π4\tan \dfrac{\pi }{4} in place of 1,
⇒tan⁡−1x+tan⁡−1y=nπ−tan⁡−1tan⁡π4\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = n\pi - {\tan ^{ - 1}}\tan \dfrac{\pi }{4}
Cancel out tan⁡−1{\tan ^{ - 1}} and tan⁡\tan we get,
⇒tan⁡−1x+tan⁡−1y=nπ−π4\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = n\pi - \dfrac{\pi }{4}
Since the option is in interval 1st and 2nd quadrant. So, substitute the value of n equal to 1,
⇒tan⁡−1x+tan⁡−1y=π−π4\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi - \dfrac{\pi }{4}
Take LCM on the left side,
⇒tan⁡−1x+tan⁡−1y=4π−π4\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \dfrac{{4\pi - \pi }}{4}
Subtract the values in the numerator,
⇒tan⁡−1x+tan⁡−1y=3π4\Rightarrow {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \dfrac{{3\pi }}{4}

Hence, option (B) is correct.

Note: Since tan⁡x\tan x is not a one-one function, its inverse is defined only within some intervals. The principal interval for tan−1xta{n^{ - 1}}x is (−π2,π2)\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) . Whenever not explicitly mentioned, we take tan−1xta{n^{ - 1}}x in the principal interval. So, we have tan−1x∈(−π2,π2)ta{n^{ - 1}}x \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right) and tan−1y∈(−π2,π2)ta{n^{ - 1}}y \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right). So,
tan−1x+tan−1x∈(−π,π)ta{n^{ - 1}}x + ta{n^{ - 1}}x \in \left( { - \pi ,\pi } \right).
This is why we chose the value of n, so that nπ+tan⁡−1x+y1−xyn\pi + {\tan ^{ - 1}}\dfrac{{x + y}}{{1 - xy}} is within the interval (−π,π)\left( { - \pi ,\pi } \right). It can also be proved that unique n satisfies this property.