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Question

The value of tan−1(12tan2A)+tan−1(cotA)+tan−1(cot3A) for 0<A<π/4 is

A
4tan1(1)
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B
2tan1(2)
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C
0
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D
none
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Solution

The correct option is C 0
tan1(12tan2A)+tan1(cotA)+tan1(cot3A)
=tan1(tanA1tan2A)+tan1(cotA(1+cot2A)1cot4A)
=tan1(tanA1tan2A)+tan1(cotA1cot2A)
=tan1(tanA1tan2A)+tan1(tanAtan2A1)
=tan1(tanA1tan2A)tan1(tanA1tan2A)
=0

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