Question

If $\cos \theta =\frac{3}{5},$ then the value of $\frac{\sin \theta -\tan \theta +1}{2{\tan }^2\theta }$

• A. $\frac{13}{15}$
• B. $\frac{91}{160}$
• C. $\frac{14}{15}$
• D. $\frac{21}{160}$

Answer 1

Answer: (D)

$\frac{21}{160}$

Let $H,B$ and $P$ be the lengths of hypotenuse, adjacent and opposite side respectively of a right triangle.

Given, $\cos \theta =\frac{3}{5}$

$\cos \theta =\frac{B}{H}=\frac{3}{5}$
By Pythagoras Theorem,
$H^2=P^2+B^2$
$5^2=P^2+3^2$
$P=4$
Now, $\frac{\sin \theta -\tan \theta +1}{2{\tan }^2\theta }$
$=\frac{\frac{P}{H}-\frac{P}{B}+1}{2(\frac{P}{B}{)}^2}$
$=\frac{\frac{4}{5}-\frac{4}{3}+1}{2(\frac{4}{3}{)}^2}$
$=\frac{\frac{12-20+15}{15}}{\frac{32}{9}}$
$=\frac{\frac{7}{15}}{\frac{32}{9}}$
$=\frac{21}{160}$