# Math Help - questions about proving of identities for trigonometry

1. ## questions about proving of identities for trigonometry

there are 2 questions.

Prove the following identities:

1.
cosec^4 Ѳ – cot^4 Ѳ = cosec^2 Ѳ + cot^2 Ѳ

2.
2 tan A = [(cos A)/(cosec A – 1)] + [(cos A)/(cosec A +1)]

for question 1, i start from the LHS, but get "stuck".
for question 2, i start from the RHS, but also get "stuck".

2. Originally Posted by wintersoltice
there are 2 questions.

Prove the following identities:

1.
cosec^4 Ѳ – cot^4 Ѳ = cosec^2 Ѳ + cot^2 Ѳ

2.
2 tan A = [(cos A)/(cosec A – 1)] + [(cos A)/(cosec A +1)]

for question 1, i start from the LHS, but get "stuck".
for question 2, i start from the RHS, but also get "stuck".
Q1: LHS factorises as a difference of two squares. One of those factors is equal to 1 from to a standard identity.

Q2: Replace cosec A with 1/sin A in RHS. Multiply top and bottom of each term by sin A to get

sin A cos A/(1 - sin A) + sin A cos A/(1 + sin A).

Get a common denominator - from a standard identity it reduces to cos^2 A. Combine the fractions etc.

3. Hello, wintersoltice!

Prove the following identities:

[1] . csc^4(θ) – cot^4(θ) .= .csc²(θ) + cot²(θ)
Factor the left side:

csc^4(θ) - cot^4(θ) . = .[csc²(θ) - cot²(θ)] [csc²(θ) + cot²(θ)]
. - . - . - . . . . . . . . . . . .\____________/
. . . . . . . . . . . . . . . . . . . . .
This is 1

Therefore, we have: .csc²(θ) + cot²(θ)

[2] . 2·tan A .= .[(cos A)/(csc A – 1)] + [(cos A)/(csc A +1)]
On the right side, combine the fractions:

. cos A - - csc A + 1 . . . . .cos A . . . csc A - 1
---------- . ----------- . + . ----------- . -----------
csc A - 1 . csc A + 1 . . . .csc A + 1 - csc A - 1

. . . . .cos A(csc A + 1) + cos A(csc A - 1)
. . = . ------------------------------------------
. . . . . . . . . .(csc A - 1)(csc A + 1)

. . . . . 2·cos A·csc A . . . . 2·cot A . . . . . .2
. . = . ----------------- . = . --------- . = . ------ . = . 2·tan A
. . . . . . csc²A - 1 . . . . . . .cot²A . . . . .cot A

4. 1.)
making into a^2-b^2=(a+b)(a-b)

left side part
cosec^4thet+cot^4theta
=(cosec^2theta+cot^2theta)(cosec^2-cot^2theta)
=cosec^2theta+cot^2theta
which is the right part

2.)

take right side part

=(cosA/coesecA-1)+(cosA/cosecA+1)
=cosA[cosecA-1+cosecA+1]/cosec^2-1
=2cosAcosecA/cot^2A
=2cotA/cot^A
=2/CotA
=2tanA which is the left part