Q6 of 120 Page 93

(tan θ +2) (2 tan θ + 1) = 5 tan θ +sec2 θ

False

L.H.S= (tan θ+2)(2 tan θ+1)


Multiplying them,


= 2 tan2 θ + tan θ + 4 tan θ + 2


= 2 tan2θ+5 tan θ+2


= 2(sec2 θ-1) +5 tan θ+2


(∵ sec2 θ - tan2 θ = 1 ⇒ tan2θ = sec2 θ-1)


= 2 sec2 θ-2+5 tan θ+2


= 5 tan θ+ 2 sec2 θ ≠R.H.S


∴, L.H.S ≠ R.H.S


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