# If 16 cot x = 12, then sin x−cos xsin x+cos x equals

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

Solution:

We are given $16 \cot x=12$. We are asked to find the following

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

We know that: $\cot x=\frac{\text { Base }}{\text { Perpendicular }}$

$\Rightarrow$ Base $=3$

$\Rightarrow$ Perpendicular $=4$

$\Rightarrow$ Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$

$\Rightarrow$ Hypotenuse $=\sqrt{16+9}$

$\Rightarrow$ Hypotenuse $=5$

Now we have

$16 \cot x=12$

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

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

We know $\sin x=\frac{\text { Perpendicular }}{\text { Hypotenuse }}$ and $\cos x=\frac{\text { Base }}{\text { Hypotenuse }}$

Now we find

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

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

$=\frac{\frac{1}{5}}{\frac{7}{5}}$

$=\frac{1}{7}$

Hence the correct option is $(a)$