Prove that 2 sec 2 A – sec 4 A – 2 cosec 2 A + cosec 4 A = cot 4 A – tan 4 A

Question

Aditya T · Answer

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 · Answer

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

Kushagra Madhukar · Answer

Dear student, Please find the solution to your problem. 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

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Prove that 2 sec 2 A – sec 4 A – 2 cosec 2 A + cosec 4 A = cot 4 A – tan 4 A

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

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Prove that 2 sec 2 A – sec 4 A – 2 cosec 2 A + cosec 4 A = cot 4 A – tan 4 A

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

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