Question

2sec²θ–sec⁴θ–2cosec²θ+cosec⁴θ = cor⁴θ–tan⁴θ,Prove it by using trigonometric identities.

Answer 1

SOLUTION :

Given :

2sec²θ–sec⁴θ–2cosec²θ+cosec⁴θ = cot⁴θ–tan⁴θ

L.H.S = 2sec²θ–sec⁴θ–2cosec²θ+cosec⁴θ

= (cosec⁴θ - 2cosec²θ) - ( sec⁴θ - 2sec²θ)

= (cosec⁴θ - 2cosec²θ + 1) - (sec⁴θ - 2sec²θ +1)

= ((cosec²θ)² - 2cosec²θ + 1²) - (( sec²θ )² - 2sec²θ + 1²)

=( cosec²θ - 1)² - (sec²θ -1)²

[ a² + b² - 2ab = (a - b)²]

= (cot²θ)² - (tan²θ)²

[ cosec²θ- 1= cot²θ , sec²θ - 1 = tan²θ]

LHS = cot⁴θ - tan⁴θ = RHS

HOPE THIS ANSWER WILL HELP YOU..

Answer 2

Here I am using A instead of theta .

LHS

= 2sec²A-sec⁴A-2cosec²A+cosec⁴A

Rearranging the terms ,we get

= (cosec⁴A-sec⁴A)-2(cosec²A-sec²A)

=[(cosec² A)²-(sec²A)²]

-2(cosec² A-sec²A)

= (cosec² A+sec ²A)(cosec²A-sec²A)

-2(cosec²A-sec²A)

[ By algebraic identity ,

( a + b )( a - b ) = a² - b² ]

=(cosec² A - sec²A)[cosec² A+sec²A-2]

=[(1+cot²A)-(1+tan²A)]

[( 1 + cot²A )+( 1 + tan²A ) - 2 ]

[ By Trigonometric identities,

i ) cosec² A = 1 + cot²A,

ii ) sec²A = 1 + tan²A ]



= [ 1 + cot²A - 1 - tan²A ]

[ 1 + cot²A + 1 + tan²A - 2 ]

= ( cot²A - tan²A )( cot²A + tan²A )

= ( cot²A )² - ( tan²A )²

= cot⁴A - tan⁴A

= RHS

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