3. find the values of the other five trigonometric functions if it is known that tanθ = 1 and π < θ < 3π/2

Explanation:

Step1: Recall the identity $\tan\theta=\frac{\sin\theta}{\cos\theta}$

Since $\tan\theta = 1$, we have $\frac{\sin\theta}{\cos\theta}=1$, so $\sin\theta=\cos\theta$.

Step2: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$

Substitute $\sin\theta=\cos\theta$ into the identity: $\sin^{2}\theta+\sin^{2}\theta=1$, $2\sin^{2}\theta = 1$, $\sin^{2}\theta=\frac{1}{2}$, $\sin\theta=\pm\frac{\sqrt{2}}{2}$.

Step3: Determine the sign of $\sin\theta$ and $\cos\theta$

Given $\pi<\theta<\frac{3\pi}{2}$, in this quadrant, both $\sin\theta$ and $\cos\theta$ are negative. So $\sin\theta=-\frac{\sqrt{2}}{2}$ and $\cos\theta=-\frac{\sqrt{2}}{2}$.

Step4: Find $\csc\theta$

$\csc\theta=\frac{1}{\sin\theta}=\frac{1}{-\frac{\sqrt{2}}{2}}=-\sqrt{2}$.

Step5: Find $\sec\theta$

$\sec\theta=\frac{1}{\cos\theta}=\frac{1}{-\frac{\sqrt{2}}{2}}=-\sqrt{2}$.

Step6: Find $\cot\theta$

$\cot\theta=\frac{1}{\tan\theta}=\frac{1}{1} = 1$.

Answer:

$\sin\theta=-\frac{\sqrt{2}}{2}$, $\cos\theta=-\frac{\sqrt{2}}{2}$, $\csc\theta=-\sqrt{2}$, $\sec\theta=-\sqrt{2}$, $\cot\theta = 1$