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rigliztetbf

rigliztetbf

Answered question

2022-06-26

If sec 8 θ a + tan 8 θ b = 1 a + b ,, Then prove that a b 0

Answer & Explanation

Layla Love

Layla Love

Beginner2022-06-27Added 29 answers

Trivially, sec 8 θ 1 and tan 8 θ 0 for all θ R for which sec θ and tan θ are defined.
If a>0 and b>0. Then, sec 8 θ a + tan 8 θ b 1 a + 0 b = 1 a > 1 a + b , a contradiction.
If a<0 and b<0. Then, sec 8 θ a + tan 8 θ b 1 a + 0 b = 1 a < 1 a + b , a contradiction.
Thus, either a 0 and b 0 or a 0 and b 0. Therefore, a b 0

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