Solveeit Logo
Login

Question

Question: Find the value of \(x + y + z\) if \({\cot ^{ - 1}}x + {\cot ^{ - 1}}y + {\cot ^{ - 1}}z = \dfrac{\p...

Find the value of x+y+zx + y + z if cot⁡−1x+cot⁡−1y+cot⁡−1z=π2{\cot ^{ - 1}}x + {\cot ^{ - 1}}y + {\cot ^{ - 1}}z = \dfrac{\pi }{2}
A) xyzxyz
B) xy+yz+zxxy + yz + zx
C) 2xyz2xyz
D) None of these

Explanation

Solution

Hint- First, we will find the value of cot⁡−1x{\cot ^{ - 1}}x in order to make our solution simple. Then we will replace that value in the given equation i.e. cot⁡−1x+cot⁡−1y+cot⁡−1z=π2{\cot ^{ - 1}}x + {\cot ^{ - 1}}y + {\cot ^{ - 1}}z = \dfrac{\pi }{2}. We will also use the property of (tan⁡(π−θ)=−tan⁡(θ)\tan( \pi - \theta) = -\tan(\theta) ).

Complete step-by-step answer:
The equation given to us by the question is: cot⁡−1x+cot⁡−1y+cot⁡−1z=π2{\cot ^{ - 1}}x + {\cot ^{ - 1}}y + {\cot ^{ - 1}}z = \dfrac{\pi }{2}.
Now, as we all know that cot⁡−1x+tan⁡−1x=π2{\cot ^{ - 1}}x + {\tan ^{ - 1}}x = \dfrac{\pi }{2}, we will find out the value of cot⁡−1x{\cot ^{ - 1}}x i.e. cot⁡−1x=π2−tan⁡−1x{\cot ^{ - 1}}x = \dfrac{\pi }{2} - {\tan ^{ - 1}}x. Now, we will replace this value of cot⁡−1x{\cot ^{ - 1}}x, in the equation given to us, we get-
→π2−tan⁡−1x+π2−tan⁡−1y+π2−tan⁡−1z=π2\to \dfrac{\pi }{2} - {\tan ^{ - 1}}x + \dfrac{\pi }{2} - {\tan ^{ - 1}}y + \dfrac{\pi }{2} - {\tan ^{ - 1}}z = \dfrac{\pi }{2}
Cancelling π2\dfrac{\pi }{2} from the above equation we will have simplified equation as-
→tan⁡−1x+tan⁡−1y+tan⁡−1z=π\to {\tan ^{ - 1}}x + {\tan ^{ - 1}}y + {\tan ^{ - 1}}z = \pi
Taking tan⁡−1z{\tan ^{ - 1}}z to the right side of the equation we get-
→tan⁡−1x+tan⁡−1y=π−tan⁡−1z\to {\tan ^{ - 1}}x + {\tan ^{ - 1}}y = \pi - {\tan ^{ - 1}}z
Multiplying the above equation by tan⁡\tan , we will get-
→tan⁡(tan⁡−1x+tan⁡−1y)=tan⁡(π−tan⁡−1z)\to \tan \left( {{{\tan }^{ - 1}}x + {{\tan }^{ - 1}}y} \right) = \tan \left( {\pi - {{\tan }^{ - 1}}z} \right)
By applying the property of tan⁡(π−θ)=−tan⁡θ\tan \left( {\pi - \theta } \right) = - \tan \theta and the property of tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) in the above equation we will get-
→tan⁡[tan⁡−1(x+y1−xy)]=−tan⁡(tan⁡−1z)  ⇒tan⁡[tan⁡−1(x+y1−xy)]=−z  ⇒x+y1−xy=−z  ⇒x+y=−z+xyz  ⇒x+y+z=xyz  \to \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)} \right] = - \tan \left( {{{\tan }^{ - 1}}z} \right) \\\ \\\ \Rightarrow \tan \left[ {{{\tan }^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)} \right] = - z \\\ \\\ \Rightarrow \dfrac{{x + y}}{{1 - xy}} = - z \\\ \\\ \Rightarrow x + y = - z + xyz \\\ \\\ \Rightarrow x + y + z = xyz \\\
Thus, we get the value of x+y+zx + y + z to be xyzxyz
Hence, above option A is correct.

Note: Don’t forget to use the properties. Replacing some values by using properties will make the answer simple. The properties like tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy){\tan ^{ - 1}}x + {\tan ^{ - 1}}y = {\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right) and tan⁡π−θ=−tan⁡θ\tan \pi - \theta = - \tan \theta should always be in your mind.