Prove that (tanθ + 2) (2 tanθ + 1) = 5 tanθ + sec2 θ
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Answer:
(tanθ + 2) (2 tanθ + 1) = 5 tanθ + sec2 θ
Step-by-step explanation:
l.H.S = (tanθ + 2) (2 tanθ + 1)
= 2 tan2 A + tanA + 4 tanA +2
=2(1- tan2A) +5tanA +2
= 2 sec2A + 5 tan A + 2
2 is missing in question
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