Question

If 3 cot θ = 4, find the value of$\frac{4cos\Theta-sin\Theta}{2cos\Theta+sin\Theta}$.

Answer 1

SOLUTION :  

Given : 3 cot θ = 4  

We have to find the value of :  (4 cos θ - sin θ)/(2cos θ + sin θ)

In right angle ∆,  

cot θ = Base / perpendicular

cos θ  = Base/ hypotenuse

sin θ = perpendicular/hypotenuse

3 cot θ = 4  

cot θ = 4 /3 = Base / perpendicular

Hypotenuse = √( perpendicular)² + (Base)²

[By Pythagoras theorem]

Hypotenuse = √ 3² + 4² = √9 + 16 = √25

Hypotenuse = √25 = 5  

Hypotenuse = 5  

cos θ  = Base/ hypotenuse = ⅘  

cos θ  = ⅘

sin θ = perpendicular/hypotenuse = ⅗

sin θ = 3/5

The value of : (4 cos θ - sin θ)/(2cos θ + sin θ)

= (4 × ⅘ - ⅗) /(2 × ⅘ + ⅗)

= (16/5 - ⅗) / (8/5 + ⅗)

= (16 - 3)/5 / (8 + 3)/5

= (13/5) / (11/5)

= 13/5 × 5/11

(4 cos θ - sin θ)/(2cos θ + sin θ) = 13/11

Hence, the value of (4 cos θ - sin θ)/(2cos θ + sin θ) is 13/11 .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

cot x =4/3
i.e cos x /sin x = 4/3
cos x = 4/3 sin x
substituting cos x = 4/3 sin x in given equation we get ,
(16/3 sin x - sin x )/(8/3 sin x + sin x )
=(13/3 sin x )/(11/3 sin x )
=13/11