Question

If $\sin \Theta =\frac{3}{5}$ and $\Theta$ is acute then find the value of $\frac{\tan \Theta -2\cos \Theta }{3\sin \Theta +\sec \Theta }$

• A. $\frac{15}{61}$
• B. $\frac{17}{61}$
• C. $-\frac{15}{61}$
• D. $-\frac{17}{61}$

Answer 1

The correct option is D $-\frac{17}{61}$

If $\sin \Theta =\frac{3}{5}$
Then, opposite side of the right angled triangle containing angle $\theta =3$ and hypotenuse $=5$
Using pythagoras theorem, we get the adjacent side $=4$

Now we have, $\cos \Theta =\frac{Adjacent}{Hypotenuse}=\frac{4}{5}$
And $\tan \Theta =\frac{Opposite}{Adjacent}=\frac{3}{4}$
So, $\frac{\tan \Theta -2cos\Theta }{3\sin \Theta +\sec \Theta }=\frac{\frac{3}{4}-2\left(\frac{4}{5}\right)}{3\left(\frac{3}{5}\right)+\frac{5}{4}}=-\frac{17}{61}$