Question

Find all solutions to the equation sec2β−1/tan(−β) =1 for which−π≤β<π.

Answer 1

The solutions to the given equation within the given range are β = -π/8, π/8, 7π/8, -7π/8.

Step-by-step explanation:

The given equation is sec^2(β) - 1/tan(-β) = 1. To find all solutions within the given range, we need to simplify the equation and solve for β. First, we know that sec^2(β) can be written as 1/cos^2(β), and tan(-β) can be written as -tan(β). Substituting these values, the equation becomes 1/cos^2(β) + 1/tan(β) = 1. Multiplying through by cos^2(β) gives us 1 + cos^2(β)/sin(β) = cos^2(β). Rearranging terms and simplifying, we get sin(β) = cos(β)/sqrt(2). From here, we can use the trigonometric identity sin(β) = cos(π/2 - β) to rewrite the equation as cos(π/2 - β) = cos(β)/sqrt(2). This equation holds when either the angles inside the cosine functions are equal or the sum of the angles equals a multiple of 2π. Solving both cases, we find the solutions β = -π/8, π/8, 7π/8, -7π/8.

Answer 2

There is no solution to the equation sec2β−1/tan(−β) =1 at intervals −π ≤ β < π.

Solving trigonometry equations.

To find all the solutions to the equation sec²β - 1/tan(-β) = 1 for which -π ≤ β < π, we need to first solve the trigonometry equation first using the appropriate notations.

Given:

`sec^2 (\beta) - (1)/(tan(-\beta)) = 1`

`tan(-\beta)` can be written as
`tan(\beta)` because
`tan (-\beta) = -tan (\beta)`

`sec^2 (\beta) + (1)/(tan(\beta)) = 1`

Let write tan β in terms of sin (β) and cos (β)

`sec^2 \beta + (1)/((sin \beta)/(cos \beta))= 1`

`sec^2 \beta + \frac{cos \beta }{{sin \beta}}= 1`

Similarly, sec²β can be written in terms of sinβ and cosβ

i.e.

`(1)/(cos^2\beta)+(cos \beta)/(sin \beta)=1`

Solving the fractions by using a factor with common denominator, we have:

`(sin \beta + cos^3 \beta)/(sin \beta \ \ cos ^2 \beta) =1`

`sin \beta + cos ^3 \beta = sin \beta + cos^2 \beta`

`sin(\beta)(1+cos^2(\beta)) =sin (\beta) cos^2(\beta)`

`1+cos^2\beta=cos^2\beta`

Therefore, we can conclude that there is no solution to the intervals at −π ≤ β < π.