A geometry problem by Mohammad Khaza

$\sin x = \dfrac{AB}{AC}$

$= \dfrac{3}{5}$

$\tan x = \dfrac{AB}{BC}$

Suppose $BC = x$

Applying Pythagorean theorem

$x^2 + 3^2 = 5^2$

or, $x^2 + 9 = 25$

or, $x^2 = 25 - 9$

or, $x^2 = 16$

or, $x = \pm4$

Now,

$BC = x = \pm4$

$\tan \theta = \dfrac{3}{4}$

So $n = \tan x - \sin x$

$= \dfrac{3}{4} - \dfrac{3}{5}$

$= \dfrac{15 - 12}{20} = \dfrac{3}{20}$

sinx=3/5

or, sinx^2=9/25

or, 1-cosx^2=9/25

or,cosx^2=1-9/25

or,cosx^2=16/25

or, cosx[=4/25 [tanx>/0]

  now, tanx=sinx/cosx

      or,tanx=3/5 x 5/4

       or, tanx=3/4

so,   n=tanx-sinx

         =3/4-3/5

         =3/20