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Write ‘True’ or ‘False’ and justify your answer.(tan θ + 2) (2 tan θ + 1) = 5 tan θ + sec2 θ. |
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Answer» False Justification: 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, sec2 θ – tan2 θ = 1, we get, tan2θ = sec2 θ-1 = 2(sec2 θ-1) +5 tan θ+2 = 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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