Question
√3+i=(a+ib)(c+id),thentan−1ba+tan−1dc
has the value
has the value
Answer: Option B
:
B
√3+i=(a+ib)(c+id)
∴ac−bd=√3 and ad+bc=1
Now tan−1(ba)+tan−1(dc)
= tan−1(ab+dc1−ba.dc)= tan−1(bc+adac−bd)=tan−1(1√3)
=nπ+π6, nϵI
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:
B
√3+i=(a+ib)(c+id)
∴ac−bd=√3 and ad+bc=1
Now tan−1(ba)+tan−1(dc)
= tan−1(ab+dc1−ba.dc)= tan−1(bc+adac−bd)=tan−1(1√3)
=nπ+π6, nϵI
Was this answer helpful ?
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