Question

If $\cos \theta =\frac{12}{13}$ find the value of $2$ $\sin \theta -4\tan \theta$, where $\theta$ is a acute

• A. $-\frac{35}{39}$
• B. $\frac{35}{39}$
• C. $\frac{24}{39}$
• D. None of these

Answer 1

Answer: (A)

$-\frac{35}{39}$

Draw a right angled $\Delta ABC$ such that $\angle B={90}^{\circ }$ and $\angle C=\theta$ (acute angle)
$\because \cos \theta =\frac{12}{13}$ and $\cos \theta =\frac{\text{Base}}{\text{Hyp.}}=\frac{BC}{AC}$ $\Rightarrow \frac{BC}{AC}=\frac{12}{13}\Rightarrow \frac{BC}{12}=\frac{AC}{13}=k$
$\Rightarrow BC=12k,AC=13k$
$\therefore$ By Pythagoras' Theorem

$A{B}^2+B{C}^2=A{C}^2$
$\Rightarrow A{B}^2=A{C}^2-B{C}^2=(13k{)}^2-(12k{)}^2=169k^2-144k^2=25k^2$
$\Rightarrow AB=5k$
$\therefore 2\sin \theta -4\tan \theta$
$=2\times \frac{AB}{AC}-4\times \frac{AB}{BC}$
$=2\times \frac{5k}{13k}-4\times \frac{5k}{12k}=\frac{10}{13}-\frac{5}{3}$
$=\frac{30-65}{39}=-\frac{35}{39}$