Question

If 3cotθ=4, find the value of 5cosθ-2sinθ/5cosθ+3sinθ

Answer 1

We are given 3cot@ = 4
cot@ = 4/3

We know that cot@ = B/P

Then B=4 , P=3

By Pythagoras theorem

H^2 = P^2+B^2
H^2 = 3^2+4^2
H^2 = 9+16We
H^2 = 25

. .
. H = 5

We have to find
5cos@-2sin@ / 5cos@+3sin@

We know that
cos@ = B/H = 4/5
sin@ = P/H = 3/5

(5×4/5) - 2×3/5 / (5×4/5) -3×3/5

5 becomes cancel out


4-6/5 / 4-9/5

20-6/5 / 20-9/5

14/5 / 11/5

14/11 Answer