Question

If 16 $\cot x=12,$ then $\frac{\sin x-\cos x}{\sin x+\cos x}$ equals

• A. $\frac{1}{7}$
• B. $\frac{3}{7}$
• C. $\frac{2}{7}$
• D. 0

Answer 1

The correct option is A $\frac{1}{7}$

$\cot x=\frac{12}{16}$

$\cot x=\frac{3}{4}$

$\text{Also,}\cot \theta =\frac{\text{base}}{\text{perpendicular}}$

So we now construct a right triangle ABC, right angled at B such that $\angle BAC=\theta ,$ Base = 3 and perpendicular = 4

In $△$ ABC,



$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=(3{)}^2+(4{)}^2$

$A{C}^2=9+16=25$

$AC=5$

$\text{As we know}\sin \theta =\frac{\text{perpendicular}}{\text{hypotenuse}}$

$\cos \theta =\frac{\text{base}}{\text{hypotenuse}}$

$\text{Therefore,}\sin x=\frac{AB}{AC}=\frac{4}{5}$

$\cos x=\frac{BC}{AC}=\frac{3}{5}$

Now,

$\frac{\sin x-\cos x}{\sin x+\cos x}$

$=\frac{\frac{4}{5}-\frac{3}{5}}{\frac{4}{5}+\frac{3}{5}}$

$=\frac{\frac{4-3}{5}}{\frac{4+3}{5}}=\frac{1}{7}$