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  • Show that tan power 4 θ + tan power 2 θ = secpower 4 θ – sec power 2 θ.

Show that tan4θ + tan2θ = sec4θ – sec2θ.       

Answers (1)

Taking L.H.S.
tan4θ + tan2θ
(tan2θ) + tan2θ…(1)
We know that sec2θ – tan2θ = 1
Put

\tan ^{2}\theta = \sec ^{2}\theta -1    in (1)
= \left ( \sec ^{2}\theta -1 \right )^{2}+\sec ^{2}\theta -1
 = \left ( \sec ^{2}\theta \right )^{2}+\left ( 1 \right )^{2}-2\left ( \sec ^{2}\theta \right )\left ( 1 \right )+\sec ^{2}\theta -1
\left ( \because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab\right )
= \sec ^{4}\theta +1-2\sec ^{2}\theta +\sec ^{2}\theta -1
 = \sec ^{4}\theta -\sec ^{2}\theta
  LHS = RHS
Hence proved        

Posted by

infoexpert27

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