question_answer

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) \[0\]

Correct Answer: A

Solution :

[a] Let ABC be the right triangle such that \[\angle B=90{}^\circ \]and \[\angle C=x.\]

Given,\[\cot x=\frac{12}{16}=\frac{3}{4}\]

\[\therefore \,\,\,\,\,\,\cot x=\frac{BC}{AB}=\frac{3}{4}\]

Let \[BC=3k\] and \[AB=4k,\]where k is positive constant.

\[\therefore \,\,\,\,AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}\]

\[\sqrt{16{{k}^{2}}+9{{k}^{2}}}\]

\[\sqrt{25{{k}^{2}}}=5k\]

So,\[\sin x=\frac{AB}{AC}=\frac{4k}{5k}=\frac{4}{5}\] .(1)

\[\cos x=\frac{BC}{AC}=\frac{3k}{5k}=\frac{3}{5}\] .(2)

Now,  \[\frac{\sin x-\cos x}{\sin x+\cos x}=\frac{\frac{4}{5}-\frac{3}{5}}{\frac{4}{5}+\frac{3}{5}}=\frac{4-3}{4+3}=\frac{1}{7}\]

[Using eqs. (1) and (2)]