Question

\[\frac{\cot 46^\circ}{\tan 44^\circ} - 3 \frac{\sec 20^\circ}{\operatorname{cosec} 70^\circ} + 5\] is equal to ______.

Options

• −3

• 4

• 3

• −4

Answer 1

\[\frac{\cot 46^\circ}{\tan 44^\circ} - 3 \frac{\sec 20^\circ}{\operatorname{cosec} 70^\circ} + 5\] is equal to 3. 

Explanation:

Solving,

\[\Rightarrow \frac{\cot 46^\circ}{\tan 44^\circ} - 3\, \frac{\sec 20^\circ}{\operatorname{cosec} 70^\circ} + 5\]

\[\Rightarrow \frac{\cot(90^\circ - 44^\circ)}{\tan 44^\circ} - 3\, \frac{\sec(90^\circ - 70^\circ)}{\operatorname{cosec} 70^\circ} + 5\]

\[\Rightarrow \frac{\tan 44^\circ}{\tan 44^\circ} - 3\, \frac{\operatorname{cosec} 70^\circ}{\operatorname{cosec} 70^\circ} + 5\]

⇒ 1 − 3 × 1 + 5

⇒1 − 3 + 5

⇒ 3