Question

If $\cos \theta =\frac{12}{13}$, show that $\sin \theta (1-\tan \theta )=\frac{35}{156}$

Answer 1

Given,

$\cos \theta =\frac{12}{13}$

We know that,

$\cos \theta =\frac{adjacentSide}{Hypotenuse}$

Consider the figure above,

From Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

${13}^2=A{B}^2+{12}^2$

$A{B}^2=169-144=25$

$AB=5$

$\sin \theta =\frac{OppositeSide}{Hypotenuse}=\frac{5}{13}$

$\tan \theta =\frac{OppositeSide}{AdjacentSide}=\frac{5}{12}$

$\sin \theta (1-\tan \theta )=\frac{5}{13}(1-\frac{5}{12})$

$=\frac{5}{13}\left(\frac{12-5}{12}\right)$

$=\frac{5}{13}\left(\frac{7}{12}\right)$

$=\frac{35}{156}$