Question

if tan(θ) = 1 and 0° < θ < 90°, what is csc(θ)? write your answer in simplified, rationalized form. csc(θ) =

Explanation:

Step1: Recall tangent formula

$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Given $\tan(\theta) = 1$, so $\frac{\sin(\theta)}{\cos(\theta)}=1$, which means $\sin(\theta)=\cos(\theta)$.

Step2: Use Pythagorean identity

Since $\sin^{2}(\theta)+\cos^{2}(\theta)=1$ and $\sin(\theta)=\cos(\theta)$, we substitute to get $\sin^{2}(\theta)+\sin^{2}(\theta)=1$, or $2\sin^{2}(\theta)=1$. Then $\sin^{2}(\theta)=\frac{1}{2}$, and $\sin(\theta)=\frac{1}{\sqrt{2}}$ (because $0^{\circ}<\theta < 90^{\circ}$, so $\sin(\theta)>0$).

Step3: Recall cosecant formula

$\csc(\theta)=\frac{1}{\sin(\theta)}$. Substituting $\sin(\theta)=\frac{1}{\sqrt{2}}$, we get $\csc(\theta)=\sqrt{2}$.

Answer:

$\sqrt{2}$