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Question

Mathematics Question on Inverse Trigonometric Functions

If xy + yz + zx = 1, then :

A

tan⁡−1x+tan⁡−1y+tan⁡−1z=0\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = 0

B

tan⁡−1x+tan⁡−1y+tan⁡−1z=π\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \pi

C

tan⁡−1x+tan⁡−1y+tan⁡−1z=π4\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{4}

D

tan⁡−1x+tan⁡−1y+tan⁡−1z=π2\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2}

Answer

tan⁡−1x+tan⁡−1y+tan⁡−1z=π2\tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2}

Explanation

Solution

xy+yz+zx=1xy + yz + zx = 1 ......(1) Now, we know tan⁡−1x+tan⁡−1y+tan⁡−1z\tan ^{-1} x + \tan ^{-1} y + \tan ^{-1} z =tan⁡−1[x+y+z−xyz1−(xy+yz+zx)]= \tan^{-1} \left[ \frac{x+y+z-xyz}{1-\left(xy+yz+zx\right)}\right] using equation (1) we have tan⁡−1x+tan⁡−1y+tan⁡−1z\tan ^{-1} x + \tan ^{-1} y + \tan ^{-1} z =tan⁡−1(10)=tan⁡∞= \tan^{-1} \left(\frac{1}{0}\right) = \tan\infty ⇒tan⁡−1x+tan⁡−1y+tan⁡−1z=π2\Rightarrow \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \frac{\pi}{2}