Question

If 5 tan θ − 4 = 0, then the value of $\frac{5sin\Theta-4cos\Theta}{5sin\Theta+4cos\Theta}$ is
(a)$\frac{5}{3}$
(b) $\frac{5}{6}$
(c) 0
(d)$\frac{1}{6}$

Answer 1

SOLUTION :  

The correct option is (c) : 0

Given : 5 tan θ - 4 /  = 0

5 tan θ = 4  

tan θ = 4/5

In right angle ∆ ,  

tan θ =  perpendicular/base = 4/5

perpendicular = 4 , base = 5

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

[By Pythagoras theorem]

Hypotenuse = √ 4² + 5² = √16 + 25 = √41

Hypotenuse = √41

sinθ = perpendicular/hypotenuse = 4/√41

cos θ = base/ hypotenuse =  5/√41

The value of (5 sin θ - 4 cos θ  ) / (5 sin θ + 4 cos θ )  :  

=( 5 × 4/√41 - 4 × 5/√41) / ( 5 × 4/√41 + 4 × 5/√41)

= (20/√41 - 20/√41) /  (20/√41 +  20/√41)

= (0)/ (40/√41)

(5 sin θ - 4 cos θ  ) / (5 sin θ + 4 cos θ )  = 0  

Hence, the value of  (5 sin θ - 4 cos θ  ) / (5 sin θ + 4 cos θ )  is 0 .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Given that :

$5 \tan \alpha - 4 = 0 \\ \\ = > 5 \tan \alpha = 4 \\ \\ = > \tan\alpha = \frac{4}{5} \\ \\ = > \frac{ \sin\alpha }{ \cos \alpha } = \frac{4}{5} \\ \\ = > \sin \alpha = 4 \: and \: \cos \alpha = 5$

Now, we have to find the value of :

$\frac{5 \sin \alpha - 4 \cos \alpha }{5 \sin \alpha + 4 \cos \alpha } \\ \\ = > \frac{5 \times 4 - 4 \times 5}{5 \times 4 + 4 \times 5} \\ \\ = > \frac{20 - 20}{20 + 20} \\ \\ = > \frac{0}{40} \\ \\ = > 0$