Question

3 sin72°/cos 18°-- sec 32°/ cosec 58°

Answer 1

Here you go hope this helps you please mark brainliest if it helped

Answer 2

Hello, Here is You're Answer!!!



Question:-



$\implies3 \frac{sin72 \degree}{cos18 \degree} - \frac{sec32 \degree}{cosec58 \degree}$



Formula:-

Cos∅=Sin(90°-∅)



Sec∅=Cosec(90-∅)


Tan∅=Cot(90°-∅)



Method of Solution:-



$\implies3 \frac{sin72 \degree}{cos18 \degree} - \frac{sec32 \degree}{cosec58 \degree} \\ \\ \\\implies3 \frac{sin72 \degree}{cos(90 - 72 \degree)} - \frac{sec32 \degree}{cosec(90 - 32 \degree} \\ \\ \\ \\ \implies3 \frac{sin72 \degree}{sin72 \degree} - \frac{sec32 \degree}{sec32 \degree} \\ \\ \\ \\ \implies3 \frac{ \cancel{sin72 \degree}}{ \cancel{sin72 \degree}} - \frac{ \cancel{sec32 \degree}}{ \cancel{sec32 \degree}} \\ \\ \\ \\\implies3 \times 1 - 1 \\ \\ \implies3 - 1 \\ \\ \implies2$