Question

(Level-2) If A = tan 6° tan 42° and B = cot 66° cot 78, then - (A) A = 2B (B) A = -B (C) A= B (D) 3A = 2B na Tot be quo o te

JEE/Engineering Exams

Maths

Solution

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( A=tan 6^{circ} tan 42^{circ} ) and ( B=cot 66^{circ} cot 78^{circ} ) ( mathrm{A} / mathrm{B}=tan 6^{circ} tan 42^{circ} tan 66^{circ} tan 78^{circ} ) Using ( tan (60-x) tan x tan (60+x)=tan 3 x ) for ( x=18^{circ} ) we get ( tan 42^{circ} tan 18^{circ} tan 78^{circ}=tan 54^{circ} ldots .(1) ) for ( x=6^{circ} ) we get ( tan 54^{circ} tan 6^{circ} tan 66^{circ}=tan 18^{circ} ldots(2) ) Eliminating tan54 'between (1) and (2) we get ( tan 6^{circ} tan 42^{circ} tan 66^{circ} tan 78^{circ}=1 ) Hence ( mathrm{A} / mathrm{B}=1 )