Question

if tanx=4, find (5sinx-3cosx)/(5sinx+3cosx)

Answer 1

tan x = 4/1 = p / b

p = 4 , b = 1 (by the using of p and b we find h)

h^ 2 = p^2 + b^ 2

= (4) ^ 2 + (1) ^ 2

= 16 + 1

h = √17

sin x = p / h = 4/√17

cos x = b / h = 1/√17

( 5sinx - 3 cos x ) / ( 5 sin x + 3 cos x)

= ( 5 * 4/√17 - 3*1/ √17) / ( 5* 4/√17 + 3 * 1/ √17)

= ( 20/√17 - 3/√17) / ( 20/√17 + 3/√17)

= (20- 3) / √17 / (20+ 3) / √17

( we cut √17 to√17)

, we get,

20- 3 / 20 + 3

= 17 / 23

Answer 2

Answer:

17/23

Step-by-step explanation:

Given:

tanx = 4

E = (5sinx-3cosx)/(5sinx+3cosx)

(divide numerator and denominator by cosx)

= E = ( (5sinx-3cosx) /cosx ) / ( (5sinx+3cosx) /cosx )

= E = (5sinx/cosx - 3cosx/cosx) / (5sinx/cosx + 3cosx/cosx)

∵ sinx/cosx = tanx

∴ E = (5tanx - 3)/(5tanx + 3)

= E = (5(4) - 3)/(5(4) + 3)        {tanx = 4}

= E = (20 - 3)/(20 + 3)

= E = 17/23

∴ Answer = 17/23

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