A) \[\frac{\sqrt{3}}{4}\]
B) \[\frac{4}{\sqrt{3}}\]
C) \[\frac{2}{\sqrt{3}}\]
D) \[\frac{\sqrt{3}}{2}\]
Correct Answer: B
Solution :
\[\frac{1}{\cos {{290}^{o}}}+\frac{1}{\sqrt{3}\sin {{250}^{o}}}\] \[=\frac{1}{\cos {{70}^{o}}}-\frac{1}{\sqrt{3}\sin {{110}^{o}}}\] \[=\frac{\sqrt{3}\sin {{10}^{o}}-\cos {{70}^{o}}}{\sqrt{3}\sin {{110}^{o}}\cos {{70}^{o}}}\] \[=\frac{\sqrt{3}\sin ({{180}^{o}}-{{70}^{o}})-cos{{70}^{o}}}{\sqrt{3}\sin (180-{{70}^{o}})cos{{70}^{o}}}\] \[=\frac{\frac{\sqrt{3}}{2}\sin {{70}^{o}}-\frac{1}{2}\cos {{70}^{o}}}{\frac{\sqrt{3}}{2}\sin 70\cos {{70}^{o}}}\] \[\frac{\cos {{30}^{o}}\sin {{70}^{o}}-\sin {{30}^{o}}\cos {{70}^{o}}}{\frac{\sqrt{3}}{2}.\frac{1}{2}\sin {{140}^{o}}}\] \[=\frac{\sin ({{70}^{o}}-{{30}^{o}})}{\frac{\sqrt{3}}{4}\sin ({{180}^{o}}-{{40}^{o}})}\] \[=\frac{\sin {{40}^{o}}}{\frac{\sqrt{3}}{4}\sin {{40}^{o}}}=\frac{4}{\sqrt{3}}\]
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