A) \[{{\left( \frac{n}{m} \right)}^{2/3}}\]
B) \[{{\left( \frac{n}{m} \right)}^{1/3}}\]
C) \[\left( \frac{n}{m} \right)\]
D) \[{{\left( \frac{n}{m} \right)}^{2}}\]
Correct Answer: B
Solution :
\[\frac{n}{m}=\frac{\sec x-\cos x}{\operatorname{cosec}x-\sin x}=\frac{1-{{\cos }^{2}}x}{\frac{\cos x}{\frac{1-{{\sin }^{2}}x}{\sin x}}}=\frac{{{\sin }^{2}}x}{\cos x}\cdot \frac{\sin x}{{{\cos }^{2}}x}\] \[={{\left( \frac{\sin x}{\cos x} \right)}^{3}}\therefore \tan x={{\left( \frac{n}{m} \right)}^{1/3}}\]
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