A) \[1\]
B) \[-1\]
C) \[2\]
D) \[-2\]
Correct Answer: B
Solution :
\[\frac{{{\cos }^{2}}{{20}^{o}}+{{\cos }^{2}}{{70}^{o}}}{{{\sec }^{2}}{{50}^{o}}-{{\cot }^{2}}{{40}^{o}}}+2{{\operatorname{cosec}}^{2}}{{58}^{o}}-\]\[2\cot {{58}^{o}}\tan {{32}^{o}}-4\tan {{13}^{o}}\tan {{37}^{o}}\tan {{45}^{o}}\]\[\tan {{53}^{o}}\tan {{77}^{o}}\] \[=\frac{{{\cos }^{2}}{{20}^{o}}+{{\cos }^{2}}({{90}^{o}}-{{20}^{o}})}{{{\sec }^{2}}{{50}^{o}}-{{\cot }^{2}}({{90}^{o}}-{{50}^{o}})}+2\] \[{{\operatorname{cosec}}^{2}}{{58}^{o}}-2\cot {{58}^{o}}\tan ({{90}^{o}}-{{58}^{o}})\]\[-4\tan {{13}^{o}}\tan {{37}^{o}}\tan {{45}^{o}}\tan ({{90}^{o}}-{{37}^{o}})\]\[\tan ({{90}^{o}}-{{13}^{o}})\] \[=\frac{{{\cos }^{2}}{{20}^{o}}+{{\sin }^{2}}{{20}^{o}}}{{{\sec }^{2}}{{50}^{o}}-{{\tan }^{2}}{{50}^{o}}}+\]\[2{{\operatorname{cosec}}^{2}}{{58}^{o}}-2{{\cot }^{2}}{{58}^{o}}\] \[-4\tan {{13}^{o}}\tan {{37}^{o}}\tan {{45}^{o}}\cot {{37}^{o}}\cot {{13}^{o}}\] \[=\frac{1}{1}+2(cose{{c}^{2}}{{58}^{o}}-{{\cot }^{2}}{{58}^{o}})-4\]\[(\tan {{13}^{o}}\cot {{13}^{o}})(\tan {{37}^{o}}\cot {{37}^{o}})\]\[\tan {{45}^{o}}\] \[=1+2-4\times 1\times 1\times 1\] \[=3-4=-1\]
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