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Question

(C) Given that $	an ^{-1}(1-x), 	an ^{-1} x$,and $	an ^{-1}(1+x)$ are in $A$.$P$.Therefore,$2 	an ^{-1} x = 	an ^{-1}(1-x) + 	an ^{-1}(1+x)$.Usi

Vedclass · Accepted Answer

$1-x^2$

Vedclass · Answer

(C) Given that $	an ^{-1}(1-x), 	an ^{-1} x$,and $	an ^{-1}(1+x)$ are in $A$.$P$.Therefore,$2 	an ^{-1} x = 	an ^{-1}(1-x) + 	an ^{-1}(1+x)$.Using the formula $	an ^{-1} A + 	an ^{-1} B = 	an ^{-1} \left( \frac{A+B}{1-AB} ight)$,we get:$2 	an ^{-1} x = 	an ^{-1} \left( \frac{(1-x) + (1+x)}{1 - (1-x)(1+x)} ight)$.$2 	an ^{-1} x = 	an ^{-1} \left( \frac{2}{1 - (1-x^2)} ight)$.$2 	an ^{-1} x = 	an ^{-1} \left( \frac{2}{x^2} ight)$.Using the formula $2 	an ^{-1} x = 	an ^{-1} \left( \frac{2x}{1-x^2} ight)$,we have:$\frac{2x}{1-x^2} = \frac{2}{x^2}$.$x^3 = 1 - x^2$.Thus,$x^3 = 1 - x^2$.

If $0 < x < 1$ and $\tan ^{-1}(1-x), \tan ^{-1} x$ and $\tan ^{-1}(1+x)$ are in $A$.$P$.,then $x^3$ is equal to : . . . . . . .

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