VerifiedHint: Here, we have to apply the formula that $\cos \theta =\dfrac{Adjacent\text{ }side}{Hypotnuse}$. Since, the adjacent side and hypotenuse are given, we have to find the opposite side by the Pythagoras theorem, where,
${{(Hypotnuse)}^{2}}={{(Opposite\text{ }side)}^{2}}+{{(Adjacent\text{ }side)}^{2}}$. Now, we will get $\sin \theta $ and $\tan \theta $ and apply the values in $\dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta }$.
Complete step-by-step answer:
Here, we are given that $\theta $ is an acute angle such that $\cos \theta =\dfrac{3}{5}$.
Now, we have to find the value of $\dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta }$.
We know that,
$\cos \theta =\dfrac{Adjacent\text{ }side}{Hypotnuse}$
$\sin \theta =\dfrac{Opposite\text{ }side}{Hypotnuse}$
$\tan \theta =\dfrac{Opposite\text{ }side}{Adjacent\text{ }side}$
Hence, from the given data we can say that,
Adjacent side = 3
Hypotnuse = 5
Consider the following figure:
Hence, by Pythagoras theorem we have:
${{(Hypotnuse)}^{2}}={{(Opposite\text{ }side)}^{2}}+{{(Adjacent\text{ }side)}^{2}}$
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 $\dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta }$.
$\begin{align}
& \Rightarrow \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta
}=\dfrac{\dfrac{4}{5}\times \dfrac{4}{3}-1}{2\times {{\left( \dfrac{4}{3} \right)}^{2}}} \\
& \Rightarrow \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta
}=\dfrac{\dfrac{16}{15}-1}{2\times \dfrac{16}{9}} \\
& \Rightarrow \dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta
}=\dfrac{\dfrac{16}{15}-1}{\dfrac{32}{9}} \\
\end{align}$
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,
$\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$
Hence, we can write,
$\dfrac{\sin \theta \tan \theta -1}{2{{\tan }^{2}}\theta }=\dfrac{1}{15}\times \dfrac{9}{32}$
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.
