Question

Without using trigonometric table, evaluate $\frac{sin 35\textdegree}{sec 55\textdegree}+\frac{cos 35\textdegree}{cosec 55\textdegree} + sec 0\textdegree$

Answer 1

➡️Answer: 2

➡️Solution:

$\frac{sin \: 35°}{sec \: 55°} + \frac{cos \: 35°}{cosec \: 55°} + sec \: 0°$
Formula used

$sin \: (90° - \theta) = cos \:\theta \\ \\ cos \: (90° - \theta) = sin \: \theta \\ \\ sec \: \theta = \frac{1}{cos \: \theta} \\ \\ cosec \: \theta = \frac{1}{sin \: \theta}$

Now

$\frac{sin \: (90° - 55°)}{sec \: 55°} + \frac{cos \: (90° - 55°)}{cosec \: 55°} + sec \: 0° \\ \\ \frac{cos \: 55°}{sec \: 55°} + \frac{sin \: 55°}{cosec \: 55°} + 1 \\ \\ \\ \frac{cos \: 55°}{ \frac{1}{cos \: 55°}} + \frac{sin \: 55°}{ \frac{1}{sin \: 55°}} + 1 \\ \\ = {cos}^{2} 55° + {sin}^{2} 55° + 1 \\ \\ = 1 + 1 \\ \\ = 2$

Because

${sin}^{2} \theta+ {cos}^{2} \theta = 1 \\ sec \: 0° = 1$

Hope it helps you.