In Δ PQR, ∠Q = 90°, If tan R = \(\frac{1}{2}\), then what is the value of \(\frac{sec P (cos R + sin P)}{cosec R (sin R - cosec P)}\) ?
Answer & explanation
Correct answer: option 2
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 = -\(\frac{8}{3}\)