Prove that 2 sec2A – sec4A – 2 cosec2A + cosec4A = cot4A – tan4A

21 Points
8 years ago

LHS = 2 sec²A - sec⁴A - 2cosec²A + cosec⁴A

= 2 sec²A - 2cosec²A - sec⁴A + cosec⁴A

= 2(sec²A - cosec²A) -[(sec²A)² - (cosec²A)²] =2(sec²A-cosec²A)+[(sec²A-cosec²A)(sec²A+cosec²A)]

= (sec²A - cosec²A) [2 - (sec²A + cosec²A)]

= (sec²A - cosec²A) (2 - sec²A - cosec²A)

= (sec²A - cosec²A) (1 - sec²A + 1 - cosec²A)

= (sec²A - cosec²A) (tan²A - cot²A)

= (1 - tan²A - 1 - cot²A) (tan²A - cot²A)

= - (tan²A + cot²A) (tan²A - cot²A)

= - (tan⁴A + cot⁴A)

= cot⁴A - tan⁴A

= RHS

.Proved

Hamza
23 Points
6 years ago
LHS = 2sec2 θ - sec4 θ - 2cosec2 θ + cosec4 θUsing sec2 θ = tan2 θ + 1 and cosec2 θ = cot2 θ + 1, we have= 2(tan2 θ + 1) - (tan2 θ + 1)2 - 2(cot2 θ + 1) + (cot2 θ + 1)2= 2tan2 θ + 2 - tan4 θ - 1 -2tan2 θ - 2cot2 θ - 2 + cot4 θ + 1 + 2cot2 θ= cot4 θ - tan4 θ= RHS= Hence proved
3 years ago
Dear student,

LHS = 2 sec²A – sec⁴A –  2cosec²A + cosec⁴A
= 2 sec²A – 2cosec²A –  sec⁴A + cosec⁴A
= 2(sec²A – cosec²A) – [(sec²A)² –  (cosec²A)²]
=2(sec²A – cosec²A) + [(sec²A – cosec²A)(sec²A+cosec²A)]
= (sec²A – cosec²A) [2 –  (sec²A + cosec²A)]
= (sec²A – cosec²A) (2 –  sec²A –  cosec²A)
= (sec²A – cosec²A) (1 –  sec²A + 1 –  cosec²A)
= (sec²A – cosec²A) (tan²A – cot²A)
= (1 – tan²A – 1 – cot²A) (tan²A – cot²A)
= – (tan²A + cot²A) (tan²A - cot²A)
= – (tan⁴A + cot⁴A)
= cot⁴A – tan⁴A
= RHS
Hence, proved.

Thanks and regards,
Kushagra