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Question
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
Options
True
False
Solution
This statement is False.
Explanation:
L.H.S = (tan θ + 2)(2 tan θ + 1)
= 2 tan2 θ + tan θ + 4 tan θ + 2
= 2 tan2θ + 5 tan θ + 2
Since, sec2θ – tan2θ = 1, we get, tan2θ = sec2θ – 1
= 2(sec2θ – 1) + 5 tan θ + 2
= 2 sec2θ – 2 + 5 tan θ + 2
= 5 tan θ + 2 sec2 θ ≠ R.H.S
∴ L.H.S ≠ R.H.S
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Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
