If #0<= aplha,beta <= 90# and #tan(alpha+beta)=3# and #tan(alpha-beta)=2# then value of #sin(2alpha)# is?

Question

A) #1/sqrt2#
B) #-1/sqrt2#
C) #1/2#
D) None of these

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Answer 1 · 2018-08-07T05:15:07

#sin(2alpha)=1/sqrt2#

Here,

# 0 <=alpha ,beta <= 90^circ =>1^(st)Quadrant#

Let #A=alpha+beta and B=alpha-beta=>A+B=2alpha#

#:. " Using " tan(A+B)=(tanA+tanB)/(1-tanAtanB)#

#tan(2alpha)=tan[(alpha+beta)+(alpha-beta)]#

#tan(2alpha)=(tan(alpha+beta)+tan(alpha-beta))/(1- tan(alpha+beta)tan(alpha-beta))#

#tan(2alpha)=(3+2)/(1-3*2)=5/(-5)#

#tan(2alpha)=-1 <0#

Now ,

#0 <= alpha <= 90^circ=>0 <= 2alpha <= 180^circ=>1^(st) or 2^(nd)Quadrant#

But , #tan(2alpha)=-1 < 0 =>2^(nd)Quadrant#

So , #2alpha=pi-pi/4#

#:.sin(2alpha)=sin(pi-pi/4)=sin(pi/4)#

#:.sin(2alpha)=1/sqrt2#