A) \[\frac{r}{\sqrt{{{z}^{2}}+{{r}^{2}}}}\]
B) \[\frac{z}{r}\]
C) \[1\]
D) \[\frac{r}{z}\]
Correct Answer: D
Solution :
\[\cos \theta =\frac{z}{\sqrt{{{z}^{2}}+{{r}^{2}}}}\] \[\Rightarrow \,\,\,\,\,{{\cos }^{2}}\theta =\frac{{{z}^{2}}}{{{z}^{2}}+{{r}^{2}}}\,\,\,\Rightarrow \,\,{{\sin }^{2}}\theta =1-\frac{{{z}^{2}}}{{{z}^{2}}+{{r}^{2}}}\] \[\Rightarrow \,\,\,\,\,\,{{\sin }^{2}}\theta =\frac{{{z}^{2}}+{{r}^{2}}-{{z}^{2}}}{{{z}^{2}}+{{r}^{2}}}=\frac{{{r}^{2}}}{{{z}^{2}}+{{r}^{2}}}\] \[\Rightarrow \,\,\,\,\,\,\sin \,\theta =\sqrt{\frac{{{r}^{2}}}{{{z}^{2}}+{{r}^{2}}}}=\frac{r}{\sqrt{{{z}^{2}}+{{r}^{2}}}}\] \[\sec \theta =\frac{1}{\cos \theta }=\frac{\sqrt{{{z}^{2}}+{{r}^{2}}}}{z}\] \[\sin \theta .\sec \theta =\frac{r}{\sqrt{{{z}^{2}}+{{r}^{2}}}}\times \frac{\sqrt{{{z}^{2}}+{{r}^{2}}}}{z}=\underline{\frac{\mathbf{r}}{\mathbf{z}}}\]
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