Question

2 sec ^ 2A - sec^4A - 2 cosec ^ 2A + cosec ^4=cot ^4A - tan ^4A
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Answer 1

Magical Trignometery

To Prove → 2 sec²A - sec⁴A - 2 cosec²A + cosec⁴A = cot⁴A - tan⁴A

________________________

L.H.S → 2 sec²A - sec⁴A - 2 cosec²A + cosec⁴A

→ 2(tan²A + 1) - (tan²A + 1)² - 2(cot²A + 1) + (cot²A + 1)²

→ (tan²A + 1)(2 - tan²A - 1) - (cot²A + 1)(2 - cot²A - 1)

→ (tan²A + 1)(1 - tan²A) - (cot²A + 1)(1 - cot²A)

→ tan²A - tan⁴A + 1 - tan²A - cot²A + cot⁴A - 1 + cot²A

→ (tan²A - tan²A) + (1 - 1) + (cot²A - cot²A) + (cot⁴A - tan⁴A)

→ cot⁴A - tan⁴A = R.H.S

Hence Proved

Answer 2

$\underline{\textbf{\large{To prove}}}$

$2 \sec^{2}A - { \sec }^{4}A\: - 2{ \csc}^{2} A + { \csc }^{4} A \\ = { \cot}^{4} A - { \tan}^{4} A$

$\underline{\textbf{proof :}}$

LHS =

$2 { \sec }^{2} A - { \sec }^{4} A - 2 { \csc }^{2} A + { \csc }^{4} A$

$= 2 { \sec }^{2} A - 2 { \csc }^{2}A - { \sec}^{4} A + { \csc }^{4} A$

$= (2 { \sec}^{2} A - 2 { \csc }^{2} A) - ( { \sec }^{4} A - { \csc }^{4} A)$

$= 2( { \sec }^{2} A - { \csc }^{2}A) - ( { \sec }^{2}A + { \csc }^{2}A)( { \sec}^{2}A - { \csc }^{2}A)$

$= ( { \sec }^{2}A - { \csc }^{2}A)(2 - { \sec }^{2}A - { \csc }^{2}A)$

$= [((1 + { \tan }^{2}A) - (1 + { \cot }^{2}A))(2 - (1 + { \tan }^{2}A) - (1 + { \cot }^{2}A)]$

$= (1 + { \tan}^{2}A - 1 - { \cot}^{2} A)(2 - 1 - { \tan }^{2}A - 1 - { \cot}^{2}A)$

$= ( { \tan}^{2}A - { \cot}^{2}A)( - { \tan }^{2}A\: - { \cot }^{2}A)$

$= - ( { \tan }^{2}A - { \cot }^{2}A)( { \tan }^{2}A + { \cot }^{2}A)$

$= - ( { \tan }^{4}A - { \cot }^{4} A)$

$= - { \tan}^{4}A + { \cot }^{4}A$

$= { \cot}^{4}A - { \tan}^{2} A$

= RHS

$\underline{\textbf{Hence proved}}$