Question

( sin (A+B)=1 )
( Rightarrow sin (A+B)=sin 90^{circ} quadleft(1=sin 90^{circ}right) )
( A+B=90^{circ} )
And, ( sin (A-B)=1 / 2 )
( sin (A-B)=sin 30^{circ} quadleft(1 / 2=sin 30^{circ}right) )
( A-B=30^{circ} )
Then, ( (A+B)+(A-B)=90^{circ}+30^{circ}=120^{circ} )
( 2 A=120^{circ} Rightarrow A=60^{circ} )
Also, ( A+B=90^{circ} Rightarrow B=90^{circ}-A=90^{circ}-60^{circ}=30^{circ} )
Then, ( tan (A+2 B) cdot tan (2 A+B) )
( =tan left[60^{circ}+2left(30^{circ}right)right] cdot tan left[2left(60^{circ}right)+30^{circ}right] )
( =tan left(60^{circ}+60^{circ}right) cdot tan left(120^{circ}+30^{circ}right) )
( =tan 120^{circ} cdot tan 150^{circ} )
( =(-sqrt{3})(-1 / sqrt{3}) quadleft[tan 120^{circ}=-sqrt{3}, tan 150^{circ}=-1 / sqrt{3}right] )
( =1 quad ) (Ans.)

# If sin (a + b) = 1, sin (a - b) = then tan(a + 2B) tan(2a + B) is equal to (A) 1 (B) -1 (C) O (D) None

Solution