Given that
3 cos 80° cosec 10° + 2 sin 59° sec 31°
⇛3 cos(90°-10°) cosec 10°+2 sin (90°-31°) sec 31°
We know that
sin (90° - A) = cos A
cos (90°-A) = sin A
⇛3 sin 10° cosec 10° + 2 cos 31° sec 31°
⇛3 sin 10° (1/sin 10°) + 2 cos 31° (1/cos 31°)
⇛3 (sin 10°/sin 10°) + 2 (cos 31°/cos 31°)
⇛3 (1) + 2(1)
⇛3+2
⇛5
Answer: Therefore, 3cos 80°cosec 10°+2cos 59°sec 31° = 5.
also read similar questions: (sin teta + sec teta)^ + (cos teta+ cosec teta )^ = (1 + sec x cosec)^
https://brainly.com/question/87846?referrer
Answer:
5
Step-by-step explanation:
you can rewrite this as
3cos(90*-10*) x 1/sin(10*) + 2sin(90*-31*) x 1/cos(31*)
=3sin(10*) x 1/sin(10*) + 2cos31* x 1/cos(31*)
reduce with sin(10*) to get
3+2cos(31*) x 1/cos(31*)
cross out the cos 31* on both sides to leave you with
3+2
5
