im stuck, how do you verify the identities (make the left = the right). I have two problems

The first one:

1- tan^4x = 2sec^2x -sec^4x

The second problem:

cos(-x) / 1+sin (-x) = sec x +tan x

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im stuck, how do you verify the identities (make the left = the right). I have two problems

The first one:

1- tan^4x = 2sec^2x -sec^4x

The second problem:

cos(-x) / 1+sin (-x) = sec x +tan x

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## Answers (1)

1)

1 - tan^4(x) = 2sec^2(x) - sec^4(x)

Choose the more complex side.

RHS = 2sec^2(x) - sec^4(x)

Factor sec^2(x).

RHS = sec^2(x) ( 2 - sec^2(x) )

Use the identity sec^2(x) = tan^2(x) + 1.

RHS = sec^2(x) ( 2 - (tan^2(x) + 1) )

RHS = sec^2(x) (2 - tan^2(x) - 1)

RHS = sec^2(x) (1 - tan^2(x))

Use the identity sec^2(x) = tan^2(x) + 1 = 1 + tan^2(x)

RHS = (1 + tan^2(x))(1 - tan^2(x))

These are conjugates which multiply as a difference of squares.

RHS = 1 - tan^4(x) = LHS

2) cos(-x)/(1 + sin(-x)) = sec(x) + tan(x)

Choosing the left hand side,

LHS = cos(-x)/(1 + sin(-x))

Note that cosine is an even function, so cos(-x) = cos(x).

Sine is an odd function, so sin(-x) = -sin(x). Replacing accordingly,

LHS = cos(x)/(1 - sin(x))

Multiply top and bottom by the bottom's conjugate, 1 + sin(x). This creates a difference of squares in the denominator.

LHS = cos(x) [ 1 + sin(x) ] / (1 - sin^2(x))

Use the identity 1 - sin^2(x) = cos^2(x)

LHS = cos(x) [ 1 + sin(x) ] / cos^2(x)

The cos(x) in the numerator cancels while the cos^2(x) in the denominator reduces by 1 in power.

LHS = [1 + sin(x)] / cos(x)

Split this into two fractions.

LHS = 1/cos(x) + sin(x)/cos(x)

LHS = sec(x) + tan(x) = RHS