If theta = tan^{-1} a,phi = tan^{-1} b and ab | vedclass

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(C) Given that $	heta = 	an^{-1} a$ and $\phi = 	an^{-1} b$ with $ab = -1$.From the given equations,we have $	an 	heta = a$ and $	an \phi = b$.S

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$\frac{\pi}{2}$

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(C) Given that $	heta = 	an^{-1} a$ and $\phi = 	an^{-1} b$ with $ab = -1$.From the given equations,we have $	an 	heta = a$ and $	an \phi = b$.Substituting these into the condition $ab = -1$,we get $	an 	heta 	an \phi = -1$.This implies $	an 	heta = -\frac{1}{	an \phi} = -\cot \phi$.We know that $-\cot \phi = 	an(\phi + \frac{\pi}{2})$ or $	an(\phi - \frac{\pi}{2})$.Since the range of the principal value branch of $	an^{-1}$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$,the difference $	heta - \phi$ results in $\pm \frac{\pi}{2}$.Specifically,$	heta - \phi = 	an^{-1} a - 	an^{-1} b$. Using the identity $	an^{-1} a - 	an^{-1} b = 	an^{-1}(\frac{a-b}{1+ab})$,since $ab = -1$,the denominator becomes $0$,which corresponds to the value $\frac{\pi}{2}$ or $-\frac{\pi}{2}$.Thus,$	heta - \phi = \pm \frac{\pi}{2}$.

If $\theta = \tan^{-1} a$,$\phi = \tan^{-1} b$ and $ab = -1$,then $\theta - \phi = $

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