Question

If $\sin \theta =\frac{3}{5},$ then evaluate $\frac{\sin \theta -\frac{1}{\tan \theta }}{2\tan \theta }$

Answer 1

Given,

$\sin \theta =\frac{3}{5}$

We know that,

$\sin \theta =\frac{oppositeSide}{Hypotenuse}$

From Pythagoras theorem,

$(Hypotenuse{)}^2=(oppositeSide{)}^2+(adjacentSide{)}^2$

$5^2=3^2+(adjacentSide{)}^2$

$(adjacentSide{)}^2=25-9=16$

$\left(adjacentSide\right)=4$

$\cos \theta =\frac{AdjacentSide}{Hypotenuse}=\frac{4}{5}$

$\tan \theta =\frac{OppositeSide}{AdjacentSide}=\frac{3}{4}$

then

$\frac{\sin \theta -\frac{1}{\tan \theta }}{2\tan \theta }$

$=\frac{\frac{3}{5}-\frac{1}{\left(\frac{3}{4}\right)}}{2\left(\frac{3}{4}\right)}$

$=\frac{\frac{3}{5}-\frac{4}{3}}{\frac{3}{2}}$

$=\frac{\frac{9-20}{15}}{\frac{3}{2}}$

$=\frac{-11}{15}\times \frac{2}{3}$

$=\frac{-22}{45}$