(58) Jan 100ttan 1250 + fan 100% ta...
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(58) Jan 100ttan 1250 + fan 100% tan 1250 22 Jan (1884

JEE/Engineering Exams
Maths
Solution
132
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( cos 25^{circ}=1 ) ( tan 22^{circ} 2 tan ^{0} frac{left.100^{circ}+tan 2 pi^{circ}right)}{1-tan 100^{circ} tan 25^{circ}} )
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