In Δ PQR, ∠Q = 90°, If tan R = \(\frac{1}{2}\), then what is the value of \(\frac{sec P (cos R + sin P)}{cosec R (sin R - cosec P)}\) ?

-\(\frac{8}{3}\)

P + R = 90° ⇒ complementary angle

so sec P = cosec R

sin P = sin (90° - R) = cos R

cosec P = sec R

\(\frac{secP (cosR + cosR)}{cosecR (sinR - secR)}\)

= \(\frac{\frac{2}{\sqrt{5}} + \frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}} - \frac{\sqrt{5}}{2} }\) = \(\frac{\frac{4}{\sqrt {5}}}{\frac{2 - 5}{2\sqrt {5}}}\)

= \(\frac{4}{-3}\) × 2 = -\(\frac{8}{3}\)