Question

secΦ=13/5 show that {2sinΦ-3cosΦ/4sinΦ-9cosΦ}=3​

Answer 1

Given:

sec◇ = 13/5

Solution:

sec◇ = 1/cos◇

=> cos◇ = 5/13

cos◇ =adj/hyp =5/13

Pythagoras theorom:

Hyp^2 = (Adj side)^2 + (Opp side)^2

(13)^2 = (5)^2 + (Opp Side)^2

169=25 + (Opp side)^2

144 = (Opp side)^2

Opp side = 12

Sin◇ = opp/hyp

=> Sin◇ = 12/13

Sub sin◇ and cos◇ in equation

=> {2sin◇-3cos◇/4sin◇-9cos◇}

{2(12/13)-3(5/13) / 4(12/13)-9(5/13)}

{(24/13)-(15/13) / (48/13)-(45/13)}

{(9/13)-(3/13)}

{9/3}

=> 3

Hence, proved.