The value of tan−1(1√2)−tan−1(√5−2√61+√6)is equal to
- π6
- π4
- π3
- π12
tan−1(1√2)−tan−1(√3−√21+√6)
tan−1(1√2)−tan−1√3+tan−1√2
⟹π2−π3=π6
Another possibility is
tan−1(1√2)+tan−1√3−tan−1√2
How to solve this ?
Answer
√5−2√61+√6=√3−√21+√3⋅√2
⟹arctan√5−2√61+√6=arctan√3−arctan√2
arctan√3=π3 and
arctan√2=arccot1√2=π2−arctan1√2
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