A) \[x<y<z\]
B) \[y<z<x\]
C) \[z<x<\text{ }y\]
D) \[~x<z<\text{ }y\]
Correct Answer: A
Solution :
Since, \[x=\tan {{15}^{o}}=\tan ({{45}^{o}}-{{30}^{o}})\] \[=\frac{\tan {{45}^{o}}-\tan {{30}^{o}}}{1+\tan {{45}^{o}}\tan {{30}^{o}}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}\] \[=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{{{(\sqrt{3}-1)}^{2}}}{3-1}\] \[\frac{3+1-2\sqrt{3}}{2}=2-\sqrt{3}\] and \[y=\cos ec{{75}^{o}}\] \[\frac{1}{\sin ({{45}^{o}}+{{30}^{o}})}\] \[=\frac{1}{\sin {{45}^{o}}\cos {{30}^{o}}+\cos {{45}^{o}}\sin {{30}^{o}}}\] \[=\frac{1}{\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}}=\frac{2\sqrt{2}}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\] \[=\frac{2\sqrt{2}(\sqrt{3}-1)}{3-1}=\sqrt{6}-\sqrt{2}\] and \[z=4\sin {{18}^{o}}=4\left( \frac{\sqrt{5}-1}{4} \right)=\sqrt{5}-1\] it is clear from the above that \[(2-\sqrt{3})<(\sqrt{6}-\sqrt{2})<(\sqrt{5}-1)\] \[\Rightarrow \] \[x<y<z\]
You need to login to perform this action.
You will be redirected in
3 sec
