Question

prove that: cosec4A-2cosec2A+2sec2A-sec4A=cot4A-tan4A​

Answer 1

$LHS = cosec^{4} A - 2cosec^{2} A +2sec^{2} A - sec^{4} A$

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

$\boxed { \pink { sec^{2} A = 1 + tan^{2} A }}$

$= cosec^{2} A[ ( cot^{2} A - 1 ] + sec^{2} A [ 2 - 1 - tan^{2} A ]$

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

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

$\boxed { \orange { (a+b)(a-b) = a^{2} - b^{2} }}$

$= cot^{4} A - 1 + 1 - tan^{4} A \\= cot^{4} A - tan^{4} A\\= RHS$

Therefore.,

$\red {cosec^{4} A - 2cosec^{2} A +2sec^{2} A - sec^{4} A }$

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

•••♪

Answer 2

Answer:

${ \csc(a) }^{4} - {2 \csc(a) }^{2} + 2 { \sec(a) }^{2} - { \sec(a) }^{4} = { \cot(a) }^{4} - { \tan(a) }^{4}$

taking LHS

${ \csc(a) }^{4} - 2 { \csc(a) }^{2} + 2 { \sec(a) }^{2} - { \sec(a) }^{4}$

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

[tex]{\csc(a)}^{2}({\csc(a)}^{2}-2)

+{\sec(a)}^{2}(2-{\sec(a)}^{2})[tex]

[tex]{\csc(a)}^{2}({\csc(a)}^{2}-1-1})+{\sec(a)}^{2}(1+1-{\sec(a)}^{2})[tex]

[tex]{\csc(a)^{2}({\cot(a)}^{2}-1})+{\sec(a)}^{2}(1+{\tan(a)}^{2})[tex]