tanx-cosx
sinx/cosx-cosx=>sinx-cos^2x/cosx
we know that sin^2x+cos^2=1=>cos^2x=1-sin^2x
and also,cosx=root(1-sin^2x)
sinx-(1-sin^2x)/root(1-sin^2x)
sinx-1+sin^2x/root(1-sin^2x)
we will substitute the values of sinx=2/3
=(2/3)2+2/3-1/root(1-(2/3)^2)
=4/9+2/3-1/root(1-4/9)
=4+6-9/9/root(5)/3
=1/9*3/root(5)
=1/3root(5)
let angle A be x
sinx=opposite/hypotenuse
it means a=2, c=3 and b=√(3^2-2^2)=√5 →pyth. theorem
therefore, tanx=a/b=2/√5
cosx=b/c=√5/3
tanx-cosx=2/√5-√5/3
getting the LCM and simplifying,
=(6-5)/3√5
=1/(3√5)
or
=√5/15→when you rationalize the denominator