Question

( tan ^{-1} 2 x+tan ^{-1} 3 x=frac{pi}{4} )
( Rightarrow tan ^{-1}left(frac{2 x+3 x}{1-6 x^{2}}right)=frac{pi}{4} Rightarrow tan ^{-1}left(frac{5 x}{1-6 x^{2}}right)=frac{pi}{4} )
( Rightarrow frac{5 x}{left(1-6 x^{2}right)}=tan frac{pi}{4}=1 Rightarrow 1-6 x^{2}=5 x )
( Rightarrow 6 x^{2}+5 x-1=0 Rightarrow 6 x^{2}+6 x-x-1=0 )
( Rightarrow 6 x(x+1)-(x+1)=0 Rightarrow(x+1)(6 x-1)=0 )
( Rightarrow x=-1 quad ) or ( quad x=frac{1}{6} )
( Rightarrow x=frac{1}{6} )
( [because x=-1 mathrm{makes} text { LHS negative }] )

# Solve tan-12x + tan- 3x = -. (CBSE 2008, 09, 120, 13C]

Solution