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)] |
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