Inverse Trigonometric FunctionHard
Question
Consider two geometric progressions a1, a2, a3....... an & b1, b2, b3,.......bn with ar = 
= 2r-1 and another sequence t1, t2, t3........tn such that tr = cot-1 (2ar + br) then 
is -
= 2r-1 and another sequence t1, t2, t3........tn such that tr = cot-1 (2ar + br) then is -Options
A.0
B.π/4
C.tan-12
D.π/2
Solution
< an > is 1, 2, 22 , ..... 2n-1
< bn > is 1,
tr = cot-1 (2ar + br) = tan-1
= tan-1
= tan-1
= tan-1 2.2r-1 - tan-1 2r-1
Now
= (tan-1 2 - tan-11)
= (tan-12.2 - tan-1 2) + ....... + (tan-1 2.2n-1 - tan2n-1)
= tan-1 ∞ - tan-1 1 =
< bn > is 1,
tr = cot-1 (2ar + br) = tan-1
= tan-1
= tan-1
= tan-1 2.2r-1 - tan-1 2r-1Now
= (tan-1 2 - tan-11)= (tan-12.2 - tan-1 2) + ....... + (tan-1 2.2n-1 - tan2n-1)
= tan-1 ∞ - tan-1 1 =
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