Question

if cos theta = 4/5 then find the value of 3 cos theta + 2cosec theta /4 sin theta +2cot theta​

Answer 1

Hope this answer will be helpful!

If so, then hit thanks

Answer 2

given that cos∅= 4/5

consider any right angle ∆ ABC.

we know that cosec∅ = base/hypotenuse

therefore the hypotenuse of the ∆ABC is = 5 and base = 4

sin∅ = perpendicular/base

cot∅ = base/perpendicular

cosec∅ = hypotenuse/perpendicular

to find it's value, first we need to find the perpendicular

by Pythagoras theorem,

➡ h² = b² + p²

➡ AC² = BC² + AB²

➡ 5² = 4² + AB²

➡ 25 = 16 + AB²

➡ 25 - 16 = AB²

➡ 9 = AB²

➡ AB = √9

➡ AB = 3cm

hence, the perpendicular is 3cm.

therefore values of :-

• sin∅ = 3/4
• cot∅ = 4/3
• cosec∅ = 5/3

hence, value of (3cos∅ + 2cosec∅)/(4sin∅ + 2cot∅) is :-

$\sf = \frac{3 \times \frac{4}{5} + 2 \times \frac{5}{3} }{4 \times \frac{3}{4} + 2 \times \frac{5}{3} } \\ \\ \sf = \frac{ \frac{12}{5} + \frac{10}{3} }{ \frac{12}{4} + \frac{10}{3} } \\ \\ \sf = \frac{ \frac{(12 \times 3)}{(5 \times 3) } + \frac{(10 \times 5)}{(3 \times 5)} }{ \frac{(12 \times 3)}{(4 \times 3)} + \frac{(10 \times 4)}{(3 \times 4)} } \\ \\ \large \sf = \frac{ \frac{36}{15} + \frac{50}{15} }{ \frac{36}{12} + \frac{40}{12} } \\ \\ \sf = \frac{86}{15} \times \frac{12}{76} \\ \\ \sf = \frac{43}{5} \times \frac{4}{38} \\ \\ \sf = \frac{ \cancel{172}}{ \cancel{76}} \\ \\ \sf canceling \: both \: by \: 4 \\ \\ \sf = \boxed{ \frac{43}{19} }$