Question

Prove that (sec^4A-sec^2A)=tan^2+tan^4

Answer 1

sec⁴A-sec²A=tan²A+tan⁴A
is what we have to prove...
we can rewrite it as...
=>sec⁴A-tan⁴A=tan²A+sec²A
so now our L.H.S=sec⁴A-tan⁴A
= (sec²A)²-(tan²A)²
= {sec²A-tan²A}{sec²A+tan²A}
= 1×{sec²A+tan²A}
={sec²A+tan²A}
=RHS proved

Answer 2

LHS = sec⁴A-sec²A

LHS = (sec²A)²-(1+tan²A)

LHS = (1+tan²A)²-1-tan²A

LHS = (1)²+(tan²A)²+2(1)(tan²A)-1-tan²A

[(a+b)² = a²+b²+2ab]

LHS = 1+tan⁴A+2tan²A-1-tan²A

LHS = 1-1+2tan²A-tan²A+tan⁴A

LHS = tan²A+tan⁴A

LHS = RHS