Given tanθ=−4/3, where π/2<θ<π, what is the value of 1/(sinθ+cosθ)?

Jason asked the same question a week ago

Here is my answer

http://www.jiskha.com/display.cgi?id=1364442848

Just noticed that in the previous question

it asked for 1/sinØ + cosØ
and I did
1/sinØ + cosØ
= 1/(4/5) + (-3/5)
= 5/4 - 3/5
= 13/20

You have changed it to 1/(sinØ + cosØ)
so it would be
1/(4/5 - 3/5)
= 1/(1/5)
= 5

To find the value of 1/(sinθ+cosθ), we first need to find the values of sinθ and cosθ using the given information.

We are given tanθ = -4/3 and the restricted range of θ as π/2 < θ < π.

Since we know that tanθ = sinθ / cosθ, we can substitute the given value of tanθ into the equation:

-4/3 = sinθ / cosθ

To find the values of sinθ and cosθ, we can use the Pythagorean identity, which states that sin^2θ + cos^2θ = 1.

Let's solve for cosθ first:

cos^2θ = 1 - sin^2θ
cos^2θ = 1 - (-4/3)^2
cos^2θ = 1 - 16/9
cos^2θ = 9/9 - 16/9
cos^2θ = -7/9

Since θ lies in the second quadrant (π/2 < θ < π), the cosine value is negative. Therefore, cosθ = -√(7/9) = -√7/3.

Now, we can solve for sinθ:

-4/3 = sinθ / (-√7/3)
-4/3 = -1 * sinθ / √7
-4√7/3 = sinθ

Now that we have the values of sinθ and cosθ, we can find the value of 1/(sinθ + cosθ):

1/(sinθ + cosθ) = 1/((-4√7/3) + (-√7/3))
1/(sinθ + cosθ) = 1/(-4√7/3 - √7/3)
1/(sinθ + cosθ) = 1/(-5√7/3)
1/(sinθ + cosθ) = -3 / (5√7)

Therefore, the value of 1/(sinθ + cosθ) is -3 / (5√7).