A) \[\frac{1}{5}\]
B) \[\frac{2}{5}\]
C) \[\frac{3}{5}\]
D) \[\frac{4}{5}\]
Correct Answer: C
Solution :
[c] \[\cot \theta =\frac{8}{15}=\frac{\text{Base}}{\text{Perpendicular}}\] |
\[\therefore \] \[AC=\sqrt{{{15}^{2}}+{{8}^{2}}}\] |
\[=\sqrt{225+64}=\sqrt{289}=17\] |
\[\therefore \] \[\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}=\sqrt{\frac{1-\frac{8}{17}}{1+\frac{8}{17}}}\] |
\[=\sqrt{\frac{\frac{9}{17}}{\frac{25}{17}}}=\sqrt{\frac{9}{15}}=\frac{3}{5}\] |
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