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Question: If \(x < 0,y < 0\) such that \(xy=1\) , then write the value of \({{\tan }^{-1}}x+{{\tan }^{-1}}y\) ...

If x<0,y<0x < 0,y < 0 such that xy=1xy=1 , then write the value of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y .

Explanation

Solution

We will directly use the formula of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y which is given 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) . Thus, we will substitute the value of xy=1xy=1 and on solving we will get the answer. Further we will convert it in radian form by multiplying it with π180∘\dfrac{\pi }{180{}^\circ } . Thus, we will get an answer.

Complete step-by-step answer :
We know that inverse function of tangent function i.e. y=tan⁡−1(x)y={{\tan }^{-1}}\left( x \right) is in range of −π2-\dfrac{\pi }{2} << yy << π2\dfrac{\pi }{2} for all real numbers.
Here, we will use the formula of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y 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) . On using this we get 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)
Now, we will put value xy=1xy=1 in the above equation. We get as
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−1){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-1} \right)
tan⁡−1x+tan⁡−1y=tan⁡−1(x+y0){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{0} \right)
Now, we know that if zero is in the denominator then value becomes infinity i.e. ∞\infty . So, we can write it as
tan⁡−1x+tan⁡−1y=tan⁡−1(∞){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \infty \right)
Now, we know that tan⁡90∘=∞\tan 90{}^\circ =\infty . So, we write it as
tan⁡−1x+tan⁡−1y=tan⁡−1(tan⁡90∘){{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \tan 90{}^\circ \right)
Thus, we get as tan⁡−1x+tan⁡−1y=90∘{{\tan }^{-1}}x+{{\tan }^{-1}}y=90{}^\circ . In radian form, it is given as 90∘×π180∘=π290{}^\circ \times \dfrac{\pi }{180{}^\circ }=\dfrac{\pi }{2} .
Hence, the value of tan⁡−1x+tan⁡−1y{{\tan }^{-1}}x+{{\tan }^{-1}}y is π2\dfrac{\pi }{2} .

Note :Remember that this formula is almost similar to tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan \left( x+y \right)=\dfrac{\tan x+\tan y}{1-\tan x\tan y} . Just, here we have to find the inverse function value of the formula is little bit changed. Also, students should know that values of all angles i.e. 0∘,30∘,45∘,60∘,90∘0{}^\circ ,30{}^\circ ,45{}^\circ ,60{}^\circ ,90{}^\circ so, that it becomes easy to solve. Do not make calculation mistakes.