I want solution...please help me

sin(θ) + cos(θ) = 1

Divide both sides by √2:

1/√2 sin(θ) + 1/√2 cos(θ) = 1/√2

We do this because sin(x) = cos(x) = 1/√2 for x = π/4, and this lets us condense the left side using either of the following angle sum identities:

sin(x + y) = sin(x) cos(y) + cos(x) sin(y)

cos(x - y) = cos(x) cos(y) - sin(x) sin(y)

Depending on which identity you choose, we get either

1/√2 sin(θ) + 1/√2 cos(θ) = sin(θ + π/4)

or

1/√2 sin(θ) + 1/√2 cos(θ) = cos(θ - π/4)

Let's stick with the first equation, so that

sin(θ + π/4) = 1/√2

θ + π/4 = π/4 + 2nπ   or   θ + π/4 = 3π/4 + 2nπ

(where n is any integer)

θ = 2nπ   or   θ = π/2 + 2nπ

We get only one solution from the second solution set in the interval 0 < θ < 2π when n = 0, which gives θ = π/2.

22.83 repeating

step-by-step explanation:

137 ÷ 6 is 22.8333333 therefore we understand that 3 is repeating.

happy to

pls mark as brainliest

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.