A) \[0\]
B) \[1\]
C) \[-1\]
D) \[3\]
Correct Answer: B
Solution :
\[{{\left( \frac{\sin {{35}^{o}}}{\cos {{55}^{o}}} \right)}^{2}}+{{\left( \frac{\cos {{55}^{o}}}{\sin {{35}^{o}}} \right)}^{2}}-2\cos {{60}^{o}}\] \[={{\left\{ \frac{\sin ({{90}^{o}}-{{55}^{o}})}{\cos {{55}^{o}}} \right\}}^{2}}+\left\{ \frac{\cos ({{90}^{o}}-{{35}^{o}})}{\sin {{35}^{o}}} \right\}-2\cos {{60}^{o}}\]\[={{\left( \frac{\cos {{55}^{o}}}{\cos {{55}^{o}}} \right)}^{2}}+{{\left( \frac{\sin {{35}^{o}}}{\sin {{35}^{o}}} \right)}^{2}}-2\cos {{60}^{o}}\] \[={{1}^{2}}+{{1}^{2}}-2\times \frac{1}{2}=1\] \[\left[ \begin{align} & \because \,\,\cos ({{90}^{o}}-\theta )=\sin \theta \\ & and\,\,\sin ({{90}^{o}}-\theta )=\cos \theta \\ \end{align} \right]\]\[\left[ \because \,\,\cos {{60}^{o}}=\frac{1}{2} \right]\]
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