Evaluate: 3cos 80° cosec 10° + 2 sin 59° sec 31°
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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)^

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

7 weeks

step-by-step explanation:

convert each fraction to the least common multiple of 20, so 3/4 = 15/20 & 1/10 = 2/20, then divide 15 by 2 and you get a little over 7, so the food will last 7 weeks.