Respuesta :
Solution
[tex]\begin{gathered} \sin \theta=\frac{3}{7} \\ \text{where, } \\ \theta\text{ is in the second quadrant} \end{gathered}[/tex]- If θ is in the second quadrant, then it means that θ is between 91° and 180°.
- Also, in the second quadrant, only sine and cosec are positive. tan (θ), cot (θ), sec(θ), cos (θ) are all negative in the second quadrant.
- Let us solve the question using this knowledge:
- Cos (θ)
[tex]\begin{gathered} \sin \theta=\frac{Opposite}{\text{Hypotenuse}}=\frac{3}{7} \\ \\ \text{Opposite}^2+\text{Adjacent}^2=\text{Hypotenuse}^2 \\ 3^2+\text{Adjacent}^2=7^2 \\ 9+\text{Adjacent}^2=49 \\ \therefore\text{Adjacent}=\sqrt[]{40} \\ \\ \therefore\cos \theta=\frac{\sqrt[]{40}}{7}\text{ if }\theta\text{ is in the first quadrant} \\ \\ \cos \theta=-\frac{\sqrt[]{40}}{7}\text{ if }\theta\text{ is in the second quadrant} \end{gathered}[/tex]
- Tan (θ)
[tex]\begin{gathered} \tan \theta=\frac{\sin\theta}{\cos\theta}=\frac{3}{7}\div(-\frac{\sqrt[]{40}}{7}) \\ \\ \tan \theta=\frac{3}{7}\times\frac{7}{-\sqrt[]{40}}=-\frac{3}{\sqrt[]{40}} \end{gathered}[/tex]Cot (θ)
[tex]\begin{gathered} \cot \theta=\frac{1}{\tan \theta} \\ \\ \cot \theta=\frac{1}{-\frac{3}{\sqrt[]{40}}}=-\frac{\sqrt[]{40}}{3} \end{gathered}[/tex]Sec (θ)
[tex]\begin{gathered} \sec \theta=\frac{1}{\cos \theta} \\ \\ \sec \theta=\frac{1}{-\frac{\sqrt[]{40}}{7}}=-\frac{7}{\sqrt[]{40}} \end{gathered}[/tex]Csc (θ)
[tex]\begin{gathered} \csc \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{3}{7}} \\ \\ \therefore\csc \theta=\frac{7}{3} \end{gathered}[/tex]