Question

If $\theta$ is an acute angle such that $\cos \theta ={\displaystyle \frac{3}{5}}$, then the value of ${\displaystyle \frac{\sin \theta \tan \theta -1}{2{\tan }^2\theta }}$ is:

a)${\displaystyle \frac{16}{625}}$
b)${\displaystyle \frac{1}{36}}$
c)${\displaystyle \frac{3}{160}}$
d)${\displaystyle \frac{160}{3}}$

Answer 1

Hint: Here, we have to apply the formula that $\cos \theta ={\displaystyle \frac{Adjacentside}{Hypotnuse}}$. Since, the adjacent side and hypotenuse are given, we have to find the opposite side by the Pythagoras theorem, where,

${(Hypotnuse)}^2={(Oppositeside)}^2+{(Adjacentside)}^2$. Now, we will get $\sin \theta$ and $\tan \theta$ and apply the values in ${\displaystyle \frac{\sin \theta \tan \theta -1}{2{\tan }^2\theta }}$.

Complete step-by-step answer:

Here, we are given that $\theta$ is an acute angle such that $\cos \theta ={\displaystyle \frac{3}{5}}$.

Now, we have to find the value of ${\displaystyle \frac{\sin \theta \tan \theta -1}{2{\tan }^2\theta }}$.

We know that,

$\cos \theta ={\displaystyle \frac{Adjacentside}{Hypotnuse}}$

$\sin \theta ={\displaystyle \frac{Oppositeside}{Hypotnuse}}$

$\tan \theta ={\displaystyle \frac{Oppositeside}{Adjacentside}}$

Hence, from the given data we can say that,

Adjacent side = 3

Hypotnuse = 5

Consider the following figure:

Hence, by Pythagoras theorem we have:

${(Hypotnuse)}^2={(Oppositeside)}^2+{(Adjacentside)}^2$

From the figure we can say that,

Opposite side = AC

Adjacent side = AB

Hypotnuse = BC

Hence, we have,

${(BC)}^2={(AC)}^2+{(AB)}^2$

Now, we have to find the opposite side, AC. For that take ${(AC)}^2$ to the left side.

Now, we can write:

$\begin{align}

  & {{(AC)}^{2}}={{(BC)}^{2}}-{{(AB)}^{2}} \

 & \Rightarrow {{(AC)}^{2}}={{5}^{2}}-{{3}^{2}} \

 & \Rightarrow {{(AC)}^{2}}=25-9 \

 & \Rightarrow {{(AC)}^{2}}=16 \

\end{align}$

Next. By taking square root on both the sides we get:

$\begin{align}

  & AC=\sqrt{16} \

 & \Rightarrow AC=4 \

\end{align}$

Hence, we can write:

$\begin{align}

  & \sin \theta =\dfrac{4}{5} \

 & \tan \theta =\dfrac{4}{3} \

\end{align}$

Next, we can find ${\displaystyle \frac{\sin \theta \tan \theta -1}{2{\tan }^2\theta }}$.

$\begin{align}

  & \Rightarrow \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta

}=\dfrac{\dfrac{4}{5}\times \dfrac{4}{3}-1}{2\times {{\left( \dfrac{4}{3} \right)}^{2}}} \

 & \Rightarrow \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta

}=\dfrac{\dfrac{16}{15}-1}{2\times \dfrac{16}{9}} \

 & \Rightarrow \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta

}=\dfrac{\dfrac{16}{15}-1}{\dfrac{32}{9}} \

\end{align}$

Now, by taking LCM, we obtain:

 $\begin{align}

  & \Rightarrow \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta

}=\dfrac{\dfrac{16-15}{15}}{\dfrac{32}{9}} \

 & \Rightarrow \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta

}=\dfrac{\dfrac{1}{15}}{\dfrac{32}{9}} \

\end{align}$

We know that,

${\displaystyle \frac{{\displaystyle \frac{a}{b}}}{{\displaystyle \frac{c}{d}}}}={\displaystyle \frac{a}{b}}\times {\displaystyle \frac{d}{c}}$

Hence, we can write,

${\displaystyle \frac{\sin \theta \tan \theta -1}{2{\tan }^2\theta }}={\displaystyle \frac{1}{15}}\times {\displaystyle \frac{9}{32}}$

Now, by cancellation, we obtain:

$\begin{align}

  & \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta }=\dfrac{1}{5}\times \dfrac{3}{32} \

 & \Rightarrow \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta }=\dfrac{3}{160} \

\end{align}$

Therefore, we can say that the value of $\dfrac{\sin \theta \tan \theta -1}{2{{\tan

}^{2}}\theta }=\dfrac{3}{160}$.

So, the correct answer for this question is option (c).

Note: Here, you should be aware of the basic trigonometric formulas, especially the formulas regarding $\sin \theta ,\cos \theta$ and $\tan \theta$. If you know these three basic formulas then, you can find the other three trigonometric ratios from these three.