A) \[2+3\sqrt{2}\]
B) \[2-2\sqrt{2}\]
C) \[3-2\sqrt{2}\]
D) \[3+2\sqrt{2}\]
Correct Answer: D
Solution :
\[\sin 135{}^\circ =\sin (180{}^\circ -45{}^\circ )=\sin 45{}^\circ =\frac{1}{\sqrt{2}}\] \[\cos 120{}^\circ =\cos (180{}^\circ -60{}^\circ )=-\cos 60{}^\circ =-\frac{1}{2}\] \[\therefore \]\[\frac{\sin 135{}^\circ -\cos 120{}^\circ }{\sin 135{}^\circ +\cos 120{}^\circ }=\frac{\frac{1}{\sqrt{2}}-\left( -\frac{1}{2} \right)}{\frac{1}{\sqrt{2}}+\left( -\frac{1}{2} \right)}\] \[=\frac{\sqrt{2}+1}{\sqrt{2}-1}\times \frac{\sqrt{2}+1}{\sqrt{2}+1}\] \[=\frac{{{(\sqrt{2}+1)}^{2}}}{2-1}=2+1+2\sqrt{2}=3+2\sqrt{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
