If p, q, r are in G.P. and tan^{-1} p, tan^{- | vedclass

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(A) Given that $p, q, r$ are in $G.P.$,we have $q^2 = pr$. Also,$	an^{-1} p, 	an^{-1} q, 	an^{-1} r$ are in $A.P.$,which implies $2 	an^{-1} q = \

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$p = q = r$

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(A) Given that $p, q, r$ are in $G.P.$,we have $q^2 = pr$. Also,$	an^{-1} p, 	an^{-1} q, 	an^{-1} r$ are in $A.P.$,which implies $2 	an^{-1} q = 	an^{-1} p + 	an^{-1} r$. Using the formula $	an^{-1} x + 	an^{-1} y = 	an^{-1} \left( \frac{x+y}{1-xy} ight)$,we get $2 	an^{-1} q = 	an^{-1} \left( \frac{p+r}{1-pr} ight)$. Taking $	an$ on both sides,$\frac{2q}{1-q^2} = \frac{p+r}{1-pr}$. Since $q^2 = pr$,the denominator $1-q^2 = 1-pr$. Thus,$2q = p+r$,which means $p, q, r$ are in $A.P.$ Since $p, q, r$ are in both $A.P.$ and $G.P.$,they must be equal,i.e.,$p = q = r$.

If $p, q, r$ are in $G.P.$ and $\tan^{-1} p, \tan^{-1} q, \tan^{-1} r$ are in $A.P.$,then $p, q, r$ satisfy the relation:

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