Question

If sinθ=3/4 and θ is in quadrant II, then cos(θ)=_________;tan(θ)=__________ ;cot(θ)=________;sec(θ)=________;csc(θ)=______;Give exact values.

Answer 1

Step-by-step explanation:

sinθ = 3/4

`\begin{gathered} \sin \theta=(opposite)/(hypotenuse)\text{ = }(3)/(4) \\ \text{opposite = 3} \\ \text{hypotenuse = 4} \end{gathered}`

cos(θ) = adjacent/hypotenuse

we need to find adjacent using pythagoras' theorem:

Hypotenuse² = opposite² + adjacent²

4² = 3² + adj²

16 - 9 = adj²

7 = adj²

adjacent = √7

`\cos \mleft(\theta\mright)=\text{ }\frac{\sqrt[]{7}}{4}`
`\begin{gathered} \text{tan}\mleft(\theta\mright)=(opposite)/(adjacent) \\ \tan (\theta)\text{ = }\frac{3}{\sqrt[]{7}} \end{gathered}`
`\begin{gathered} \text{cot}(\theta)\text{ = }(1)/(\tan (\theta)) \\ \cot (\theta)\text{ = }\frac{1}{\frac{3}{\sqrt[]{7}}}\text{ = 1 }/\text{ }\frac{3}{\sqrt[]{7}}\text{ = 1}*\frac{\sqrt[]{7}}{3} \\ \cot (\theta)\text{ = }\frac{\sqrt[]{7}}{3} \end{gathered}`
`\begin{gathered} \sec (\theta)=(1)/(\cos (\theta))\text{ } \\ \sec (\theta)=\frac{1}{\frac{\sqrt[]{7}}{4}}\text{ = 1}*\text{ }\frac{\sqrt[]{7}}{4} \\ \sec (\theta)\text{ = }\frac{4}{\sqrt[]{7}} \end{gathered}`
`\begin{gathered} \csc (\theta)=\text{ }(1)/(\sin (\theta)) \\ \csc (\theta)=(1)/((3)/(4))\text{ = 1}*(4)/(3) \\ \csc (\theta)=\text{ }(4)/(3) \end{gathered}`