Question

without using trigonometric tables ? ​

Answer 1

Answer:

$\frac{tan \: 50 \degree + sec \: 50 \degree}{cot \: 40 \degree + cosec \: 40 \degree} + cos \: 40 \degree \: cosec \: 50 \degree \\ \\ = \frac{cot( 90 \degree - \: 50 \degree) + cosec \: ( 90 \degree - \: 50 \degree) \: }{cot \: 40 \degree + cosec \: 40 \degree} + cos \: 40 \degree \: sec \: ( 90 \degree - \: 50 \degree) \: \\ = \frac{cot \: 40 \degree + cosec \: 40 \degree}{cot \: 40 \degree + cosec \: 40 \degree} + cos \: 40 \degree sec \: 40 \degree \\ = 1 + cos \: 40 \degree \times \frac{1}{ cos \: 40 \degree} \: \\ = 1 + 1 \\ = 2$