Practicing Success
if θ + Ø = 90°, then \(\frac{sec Ø (1- sin Ø) ( sin Ø + cos Ø) (sec Ø + tan Ø)}{sin Ø (1+ tan Ø) + cos Ø (1 + cot Ø)}\) is equal to? |
sin θ cos θ cosec (90° - θ) sec (90° - θ) 2sin θ 2cos θ |
cosec (90° - θ) sec (90° - θ) |
\(\frac{(sec Ø - tan Ø) (sin Ø + cos Ø) (sec Ø + tan Ø)}{sin Ø (1 + \frac{sin Ø}{cos Ø}) + cos Ø ( 1 + \frac{cos Ø}{sin Ø})}\) = \(\frac{(sin Ø + cos Ø)(sec^2 Ø - tan^2 Ø) }{(sin Ø + cos Ø) (\frac{sin Ø}{cos Ø}+ \frac{cos Ø}{sin Ø})}\) = \(\frac{1}{\frac{sin^2 Ø + cos^2 Ø}{sinØ . cosØ}}\) = sin Ø cos Ø = cosec (90° - θ) sec ( 90° - θ) |