Given \tan \theta = -\frac{4}{3}, where \frac{\pi}{2} < \theta < \pi, what is the value of \frac{1}{\sin \theta + \cos \theta}?
I will guess that you meant:
tanØ = 4/3, π/2 < Ø < π , or Ø is in quadrant II
construct the right-angled triangle, and you should recognize the 3-4-5 triangle.
guessing that you want:
1/sinØ + cosØ
in II, the sine is + , but the cosine is -
sinØ = 4/5 and cosØ = -3/5
1/sinØ + cosØ
= 1/(4/5) + (-3/5)
= 5/4 - 3/5
= 13/20
To find the value of \frac{1}{\sin \theta + \cos \theta}, we need to determine the values of \sin \theta and \cos \theta first.
We are given that \tan \theta = -\frac{4}{3}. Since \tan \theta = \frac{\sin \theta}{\cos \theta}, we can use this relationship to find the values of \sin \theta and \cos \theta.
Dividing both sides of the equation \tan \theta = -\frac{4}{3} by \cos \theta, we get:
\frac{\sin \theta}{\cos \theta} = -\frac{4}{3}
Now, remember the Pythagorean identity \sin^2 \theta + \cos^2 \theta = 1. Let's rearrange this identity to get an expression for \cos \theta:
\cos^2 \theta = 1 - \sin^2 \theta
\cos \theta = \sqrt{1 - \sin^2 \theta}
Substituting this expression for \cos \theta into the equation \frac{\sin \theta}{\cos \theta} = -\frac{4}{3}, we can solve for \sin \theta:
\frac{\sin \theta}{\sqrt{1 - \sin^2 \theta}} = -\frac{4}{3}
Squaring both sides of the equation, we get:
\frac{\sin^2 \theta}{1 - \sin^2 \theta} = \left(-\frac{4}{3}\right)^2
\frac{\sin^2 \theta}{1 - \sin^2 \theta} = \frac{16}{9}
Cross-multiplying, we have:
\sin^2 \theta = \frac{16}{9} - \frac{16}{9}\sin^2 \theta
\sin^2 \theta + \frac{16}{9}\sin^2 \theta = \frac{16}{9}
Combining like terms, we get:
\left(1 + \frac{16}{9}\right)\sin^2 \theta = \frac{16}{9}
\left(\frac{25}{9}\right)\sin^2 \theta = \frac{16}{9}
Dividing both sides of the equation by \frac{25}{9}, we obtain:
\sin^2 \theta = \frac{16}{25}
\sin \theta = \pm \frac{4}{5}
Since \frac{\pi}{2} < \theta < \pi, where the cosine function is negative, we have:
\cos \theta = -\sqrt{1 - \sin^2 \theta}
\cos \theta = -\sqrt{1 - \left(\frac{4}{5}\right)^2}
\cos \theta = -\frac{3}{5}
Now, we can find the value of \frac{1}{\sin \theta + \cos \theta}:
\frac{1}{\sin \theta + \cos \theta} = \frac{1}{\left(\frac{4}{5}\right) + \left(-\frac{3}{5}\right)}
\frac{1}{\sin \theta + \cos \theta} = \frac{1}{\frac{1}{5}}
\frac{1}{\sin \theta + \cos \theta} = \boxed{5}
To find the value of \(\frac{1}{\sin \theta + \cos \theta}\), we first need to find the values of \(\sin \theta\) and \(\cos \theta\).
Given that \(\tan \theta = -\frac{4}{3}\), we can use the identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) to find the values of \(\sin \theta\) and \(\cos \theta\).
\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)
\(-\frac{4}{3} = \frac{\sin \theta}{\cos \theta}\)
Next, we can use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\) to find another equation involving \(\sin \theta\) and \(\cos \theta\).
\(\sin^2 \theta + \cos^2 \theta = 1\)
Now we have two equations:
1) \(-\frac{4}{3} = \frac{\sin \theta}{\cos \theta}\)
2) \(\sin^2 \theta + \cos^2 \theta = 1\)
To solve these equations, we can use substitution. Rearrange equation 1) to solve for \(\cos \theta\):
\(\cos \theta = \frac{\sin \theta}{-\frac{4}{3}} = -\frac{3}{4} \sin \theta\)
Substitute this expression for \(\cos \theta\) in equation 2):
\(\sin^2 \theta + \left( -\frac{3}{4} \sin \theta \right)^2 = 1\)
Simplify:
\(\sin^2 \theta + \frac{9}{16} \sin^2 \theta = 1\)
Combine like terms:
\(\frac{25}{16} \sin^2 \theta = 1\)
Divide by \(\frac{25}{16}\):
\(\sin^2 \theta = \frac{16}{25}\)
Take the square root of both sides:
\(\sin \theta = \pm \frac{4}{5}\)
Since \(\frac{\pi}{2} < \theta < \pi\), \(\sin \theta\) is negative in this range. So:
\(\sin \theta = -\frac{4}{5}\)
Substitute this value into equation 1) to find \(\cos \theta\):
\(-\frac{4}{3} = \frac{-\frac{4}{5}}{\cos \theta}\)
Cross multiply:
\(-4 \cos \theta = -\frac{12}{5}\)
Divide by -4:
\(\cos \theta = \frac{3}{5}\)
Now we have the values of \(\sin \theta = -\frac{4}{5}\) and \(\cos \theta = \frac{3}{5}\).
To find the value of \(\frac{1}{\sin \theta + \cos \theta}\), substitute these values:
\(\frac{1}{\sin \theta + \cos \theta} = \frac{1}{\left(-\frac{4}{5}\right) + \frac{3}{5}}\)
Simplify:
\(\frac{1}{\frac{-4}{5} + \frac{3}{5}} = \frac{1}{\frac{-1}{5}}\)
Invert the denominator:
\(\frac{1}{\frac{-1}{5}} = \frac{5}{-1} = -5\)
Therefore, the value of \(\frac{1}{\sin \theta + \cos \theta}\) is \(-5\).