Question : If $\cos 48^{\circ}=\frac{m}{n}$, then $\sec 48^{\circ}-\cot 42^{\circ}$ is equal to:
Option 1: $\frac{m-\sqrt{n^2-m^2}}{m}$
Option 2: $\frac{m-\sqrt{n^2-m^2}}{n}$
Option 3: $\frac{n-\sqrt{n^2-m^2}}{n}$
Option 4: $\frac{n-\sqrt{n^2-m^2}}{m}$
Correct Answer: $\frac{n-\sqrt{n^2-m^2}}{m}$
Solution : Given that $\cos 48^{\circ}=\frac{m}{n}$, We know that $\sec 48^{\circ}=\frac{n}{m}$, We know that $\tan \theta=\sqrt{\sec^2 \theta-1}$, Now, $\cot 42^{\circ}=\tan 48^{\circ}=\sqrt{\sec^2 48^{\circ}-1}=\sqrt{\left(\frac{n}{m}\right)^2-1}=\sqrt{\frac{n^2}{m^2}-1}=\frac{\sqrt{n^2-m^2}}{m}$ So, $\sec 48^{\circ}-\cot 42^{\circ}=\frac{n}{m}-\frac{\sqrt{n^2-m^2}}{m}=\frac{n-\sqrt{n^2-m^2}}{m}$. Hence, the correct answer is $\frac{n-\sqrt{n^2-m^2}}{m}$.
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