$$sin$$−$$1a$$−$$a23+a39+$$....$$+$$$$cos$$−$$11+b+b2+$$...$$=$$π$$2then$$

Correct option is 2a=1 and b=−13

Questions

$\sin^{- 1} (a - \frac{a^{2}}{3} + \frac{a^{3}}{9} + ....) + \cos^{- 1} (1 + b + b^{2} + .. .) = \frac{π}{2}$ then

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a=−3 and b=1

a=1 and b=−13

a=16 and b=12

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detailed solution

Correct option is B

The given relation is possible when a−a23+a39+....=1+b+b2+... Also−1≤a−a23+a39+...≤1 and −1≤1+b+b2+...≤1⇒|b|<1⇒|a|<3 and a1+a3=11−b⇒3aa+3=11−b

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sin−1a−a23+a39+....+cos−11+b+b2+...=π2then

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$\sin^{- 1} (a - \frac{a^{2}}{3} + \frac{a^{3}}{9} + ....) + \cos^{- 1} (1 + b + b^{2} + .. .) = \frac{π}{2}$ then