BRAINLIESTTT ASAP! PLEASE HELP ME :)

Work Shown:

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

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

tan^4(x)+tan^2(x) + 2tan^2(x)+2-tan^2(x) = 2

tan^4(x)+2tan^2(x)+2 = 2

tan^4(x)+2tan^2(x) = 0

tan^2(x)(tan^2(x)+2) = 0

tan^2(x) = 0 or tan^2(x)+2 = 0

tan^2(x) = 0 or tan^2(x) = -2

The equation tan^2(x) = -2 has no real solutions for similar reasons that x^2 = -2 does not have real solutions. So we ignore tan^2(x) = -2. Only focus on tan^2(x) = 0

So,

tan^2(x) = 0

tan(x) = sqrt(0)

tan(x) = 0

x = arctan(0)

x = n*pi, where n is any integer

Based on the restriction that 0 <= x < 2pi, this means the infinite set of solutions gets restricted to this subset: x = 0, x = pi, which occurs when n = 0 and n = 1 respectively.

The value of x is 0 rad and π rad

Step-by-step explanation:

Given that the basic trigonometric formula is, sec²θ = 1 + tan²θ. So you can substitute the formula into the expression :

tan²xsex²x + 2sec²x - tan²x = 2

tan²x(1+tan²x) + 2(1+tan²x) - tan²x = 2

tan²x + tan⁴x + 2 + 2tan²x - tan²x = 2

tan⁴x + 2tan²x = 0

Let tan²x = y to make an equation so that it is easier to solve :

y² + 2y = 0

y(y + 2) = 0

y = 0

y = -2

REMEMBER to convert it back :

y = 0

tan²y = 0

tan y = 0 (1st and 3rd quadrant)

(θ = 0 rad)

x = 0rad, πrad

y = -2

tan²x = -2

tanx = √-2 (rejected)

answer: i'm not sure

step-by-step explanation:

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation: