Trigonometry Problem: Calculating Tangent and Cosine
The image contains a trigonometry problem which states:
"(b) Given that \( \sin\theta = \frac{3}{5} \) and \( 90° \leq \theta \leq 180° \), find the exact value of \( \tan\theta + \cos\theta \)."
Since \( \sin\theta = \frac{3}{5} \) and the angle \( \theta \) falls in the second quadrant (because \( 90° \leq \theta \leq 180° \)), we know that the cosine of \( \theta \) will be negative, and the tangent of \( \theta \) will be negative as well because the tangent of an angle is the sine divided by the cosine, and in the second quadrant the cosine is negative.
We can find the cosine of \( \theta \) using the Pythagorean identity, where \( \cos^2\theta = 1 - \sin^2\theta \).
\( \cos^2\theta = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{25}{25} - \frac{9}{25} = \frac{16}{25} \)
Since \( \theta \) is in the second quadrant, \( \cos\theta \) will be negative:
\( \cos\theta = -\sqrt{\frac{16}{25}} = -\frac{4}{5} \)
Now we can find the tangent of \( \theta \):
\( \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4} \)
Now, we can find the value of \( \tan\theta + \cos\theta \):
\( \tan\theta + \cos\theta = -\frac{3}{4} + (-\frac{4}{5}) = -\frac{3 \cdot 5}{4 \cdot 5} - \frac{4 \cdot 4}{4 \cdot 5} = -\frac{15}{20} - \frac{16}{20} = -\frac{31}{20} \)
So, the exact value of \( \tan\theta + \cos\theta \) given \( \sin\theta = \frac{3}{5} \) and \( 90° \leq \theta \leq 180° \) is \( -\frac{31}{20} \).
