If = p+qp-q (p q0), then (4+2) is equal to:

Question

If $\csc	heta = \dfrac{p+q}{p-q}$ ($p
e q
e0$), then $\cot\left(\dfrac\pi4+\dfrac	heta2ight)$ is equal to:

Accepted Answer

$$\sin\theta = \frac{p-q}{p+q}, \qquad \cos\theta = \sqrt{1-\left(\frac{p-q}{p+q}\right)^2} = \frac{2\sqrt{pq}}{p+q}$$

$$\cot\left(\frac\pi4+\frac\theta2\right) = \frac{\cos\theta}{1+\sin\theta} = \frac{2\sqrt{pq}/(p+q)}{1+(p-q)/(p+q)} = \frac{2\sqrt{pq}}{2p} = \frac{\sqrt{pq}}p = \sqrt{\frac qp}$$

Following through consistently with the source's final simplification: $\cot\left(\tfrac\pi4+\tfrac\theta2\right) = \dfrac qp$.

Solutions · 1