Question

If 16 cot x = 12, then $\frac{sinx-cosx}{sinx+cosx}$ equals
(a)$\frac{1}{7}$
(b)$\frac{3}{7}$
(c)$\frac{2}{7}$
(d) 0

Answer 1

SOLUTION :  

The correct option is  (a)= 1/7

Given : 16 cot x = 12

cot x = 12/16 = ¾  

cot x = 3/4

cot x = Base / perpendicular = 3/4

Base = 3 , perpendicular = 4  

In right angled ∆,  

Hypotenuse² = ( perpendicular)² + (Base)²

[By Pythagoras theorem]

Hypotenuse² = ( 4)² + 3²

Hypotenuse² = 16 + 9  = 25

Hypotenuse = 25

Hypotenuse = √25 = 45

perpendicular = 5

sin x = perpendicular/hypotenuse = 4/5

sin x = ⅘  

cos x = base/ hypotenuse = 3/5

The value of : sin x - cos x /  sin x + cos x  

= (⅘ - ⅗)/(⅘ + ⅗)

= [(4 - 3)/5]  /[ (4 + 3)/5 ]

= (⅕) / (7/5)

= ⅕ × 5/7  

sin x - cos x /  sin x + cos x  = 1/7  

Hence, the value of  sin x - cos x /  sin x + cos x  is 1/7 .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

16 cot x = 12, then $\frac{sinx-cosx}{sinx+cosx}$ equals

(a)$\frac{1}{7}$

(b)$\frac{3}{7}$

(c)$\frac{2}{7}$

(d) 0