Q15 of 120 Page 95

Show that tan4 θ +tan2 θ =sec4 θ -sec2 θ.

L.H.S: tan4θ + tan2 θ

Taking tan2 θ common, we get


tan2 θ (tan2 θ + 1)


= tan2 θ sec2 θ [∵ sec2 θ – tan2 θ = 1 ⇒ tan2 θ + 1 = sec2 θ]


= (sec2 θ – 1) sec2 θ [∵tan2 θ = sec2 θ – 1]


= sec4 θ – sec2 θ


: R.H.S


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