Question

Evaluate: $\frac {3sin72\textdegree}{cos18\textdegree} - \frac {sec32\textdegree}{cosec58\textdegree}$

Answer 1

➡️Answer:2

➡️Solution:

$\frac{3 \sin(72°) }{cos \: 18°} - \frac{sec \: 32°}{cosec \: 58°}$
we know that

$sin \: (90° - \theta) = cos \: \theta \\ \\ sec(90° - \theta) = cosec \: \theta$
So
$= \frac{3 \sin(90° - 18°) }{cos \: 18°} - \frac{sec \: (90° - 58°)}{cosec \: 58°} \\ \\ because \: \: 90 - 18 = 72 \\ \\ 90 - 58 = 32 \\ \\ apply \: formula \: discussed \: above \\ \\ = \frac{3 cos \: 18° }{cos \: 18°} - \frac{cosec \: 58°}{cosec \: 58°} \\ \\ = 3 - 1 \\ \\ = 2$

Hope it helps you.