Question

tan4 θ + tan2 θ = sec4 θ - sec2 θ ​

Answer 1

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L.H.S = tan4 θ + tan2 θ

= tan2 θ (tan2 θ + 1)

= (sec2 θ - 1) (tan2 θ + 1) [since, tan2 θ = sec2 θ – 1]

= (sec2 θ - 1) sec2 θ [since, tan2 θ + 1 = sec2 θ]

= sec4 θ - sec2 θ = R.H.S.

Hence Proved

Answer 2

Answer:

$\mathfrak\red{Heya}$

Step-by-step explanation:

$Sec^{4}0-Sec^{2}  0=Tan^{4}0+Tan^{2}0$

$LHS,Sec^{4} 0-Sec^{2}0$

$Sec^{2} 0(Sec^{2} 0 -1)$

$(1+Tan^{2} 0)(1-Tan^{2} 0-1)$

$(1+Tan^{2} 0)(Tan^{2} 0)$

$Tan^{2} 0+Tan^{4} 0$

$Hence,Proved$