A) \[1+2\sqrt{3}\]
B) \[-1+2\sqrt{3}\]
C) \[1+\sqrt{3}\]
D) \[1-\sqrt{3}\]
Correct Answer: A
Solution :
We have, \[{{\left( \frac{\tan {{20}^{o}}}{\cos ec{{70}^{o}}} \right)}^{2}}+{{\left( \frac{\cot {{20}^{o}}}{\sec {{70}^{o}}} \right)}^{2}}+2\] \[\tan {{15}^{o}}\tan {{37}^{o}}\tan {{53}^{o}}\tan {{60}^{o}}\tan {{75}^{o}}\] \[={{\left( \frac{\sin {{20}^{o}}}{\cos {{20}^{o}}}\times \sin {{70}^{o}} \right)}^{2}}+{{\left( \frac{\cos {{20}^{o}}}{\sin {{20}^{o}}}\times \cos {{70}^{o}} \right)}^{2}}\] \[+2\tan {{15}^{o}}\tan {{37}^{o}}\tan {{(90-37)}^{o}}\] \[(\tan {{60}^{o}}\tan {{(0-15)}^{o}}\] \[[\therefore \,\,\,\tan {{(90-\theta )}^{o}}=\cot \theta ]\] \[={{\left[ \frac{\sin {{20}^{o}}}{\cos {{20}^{o}}}\times \sin {{(90-20)}^{o}} \right]}^{2}}+\] \[{{\left[ \frac{\cos {{20}^{o}}}{\sin {{20}^{o}}}\times \cos {{(90-20)}^{o}} \right]}^{2}}\] \[+2tsn{{15}^{o}}\tan {{37}^{o}}\cot {{37}^{o}}\tan {{60}^{o}}\cot {{15}^{o}}\] \[={{(\sin {{20}^{o}})}^{2}}+{{(\cos {{20}^{o}})}^{2}}+2\tan {{60}^{o}}\] \[=1+2\sqrt{3}\] \[[\because \,\,{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1]\]
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