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1 solution
Since, t a n − 1 x = x − ( x 3 / 3 ) + ( x 5 / 5 ) . . . for ∣ x ∣ ≤ 1 = > t a n − 1 1 = 1 − ( 1 / 3 ) + ( 1 / 5 ) . . . But, t a n − 1 1 = 4 π = > 1 − ( 1 / 3 ) + ( 1 / 5 ) . . . = 4 π Hence, = > ( 4 π ) / π = 4 1 = 0 . 2 5