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If cot^(-1)x+cot^(-1)y+cot^(-1)z=pi/2,x ...

If `cot^(-1)x+cot^(-1)y+cot^(-1)z=pi/2,x , y , z >0a n dx y<1,` then `x+y+z` is also equal to `1/x+1/y+1/z` (b) `x y z` `x y+y z+z x` (d) none of these

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To solve the problem, we start with the equation given in the question: \[ \cot^{-1} x + \cot^{-1} y + \cot^{-1} z = \frac{\pi}{2} \] ### Step 1: Rewrite the equation using the identity for cotangent Using the identity that states \(\cot^{-1} a = \frac{\pi}{2} - \tan^{-1} a\), we can rewrite the equation: ...
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