Given that,
PR + QR = 25
PQ = 5. Let PR be represented by x.
Therefore, QR = 25 - x.

Applying the Pythagorean theorem to ΔPQR, we get:
\(\text{PR}^ 2 = \text{PQ}^ 2 + \text{QR}^ 2 \)
\(x^2= (5)^ 2 + (25 - x)^ 2 \)
\(x^2= 25 + 625 + x^ 2 - 50x \)
\(50x=650\)
\(x=13\)
Thus, PR = 13 cm.
QR = (25 - 13) cm = 12 cm.
\(\text{sin p} =\frac{\text{ Opposite Side}}{\text{Hypotenuse }}= \frac{QR}{PR} = \frac{12}{13}\)
\(\text{sin p} = \frac{\text{Opposite Side}}{\text{Hypotenuse }}= \frac{QR}{PR} =\frac{ 12}{13}\)
\(\text{tan p} =\frac{\text{Opposite Side}}{\text{Adjacent side }}= \frac{QR}{PQ} = \frac{12}{5}\)
If \(cot\ \theta = \frac{7}{8},\) evaluate:
(i) \(\frac{(1 + sin\ \theta)(1 – sin θ)}{(1+cos θ)(1-cos θ)}\)
(ii) \(cot^2\) \(θ\)