The value of tan−1(12tan2A)+tan−1(cotA)+tan−1(cot3A) for 0<A<π/4 is
A
4tan−1(1)
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B
2tan−1(2)
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C
0
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D
none
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Solution
The correct option is C 0 tan−1(12tan2A)+tan−1(cotA)+tan−1(cot3A) =tan−1(tanA1−tan2A)+tan−1(cotA(1+cot2A)1−cot4A) =tan−1(tanA1−tan2A)+tan−1(cotA1−cot2A) =tan−1(tanA1−tan2A)+tan−1(tanAtan2A−1) =tan−1(tanA1−tan2A)−tan−1(tanA1−tan2A) =0