Question

If $\sin \theta =\dfrac{12}{13}$, then the value of $\dfrac{2\cos \theta +3\tan \theta }{\sin \theta +\tan \theta \sin \theta }$ is
a. $\dfrac{12}{5}$
b. $\dfrac{5}{13}$
c. $\dfrac{259}{102}$
d. $\dfrac{259}{65}$

Answer 1

Hint: In order to solve this question, we should know the relation of trigonometric ratios like, if $\sin \theta =\dfrac{a}{b}$ then $\cos \theta =\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}$ and $\tan \theta =\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}$. By using these properties, we will be able to find the value of the given expression.

Complete step-by-step answer:

In this question, we have been asked to find the value of $\dfrac{2\cos \theta +3\tan \theta }{\sin \theta +\tan \theta \sin \theta }$ when it is given that $\sin \theta =\dfrac{12}{13}$.To solve this question, we should know the relation between trigonometric angles like, if $\sin \theta =\dfrac{a}{b}$ then $\cos \theta =\dfrac{\sqrt{{{b}^{2}}-{{a}^{2}}}}{b}$ and $\tan \theta =\dfrac{a}{\sqrt{{{b}^{2}}-{{a}^{2}}}}$. Now, we have been given that $\sin \theta =\dfrac{12}{13}$. So, for a = 12 and b = 13, we can write, $\cos \theta =\dfrac{\sqrt{{{13}^{2}}-{{12}^{2}}}}{13}$ and $\tan \theta =\dfrac{12}{\sqrt{{{13}^{2}}-{{12}^{2}}}}$.
And we can further write them as,
$\cos \theta =\dfrac{\sqrt{169-144}}{13}$ and $\tan \theta =\dfrac{12}{\sqrt{169-144}}$
$\cos \theta =\dfrac{\sqrt{25}}{13}$ and $\tan \theta =\dfrac{12}{\sqrt{25}}$.
$\cos \theta =\dfrac{5}{13}$ and $\tan \theta =\dfrac{12}{5}$.
Now, we will put the value of $\sin \theta ,\cos \theta $ and $\tan \theta $ in the given expression, that is $\dfrac{2\cos \theta +3\tan \theta }{\sin \theta +\tan \theta \sin \theta }$. So, we will get,
$\dfrac{2\times \dfrac{5}{13}+3\times \dfrac{12}{5}}{\dfrac{12}{13}+\dfrac{12}{5}\times \dfrac{12}{13}}$
Now, we will simplify it further, so we get,
$\dfrac{\dfrac{10}{13}+\dfrac{36}{5}}{\dfrac{12}{13}+\dfrac{144}{65}}$
Now, we will take the LCM of both the terms of the numerator and denominator. So, we will get,
\[\begin{align}
  & \dfrac{\dfrac{10\times 5+36\times 13}{13\times 5}}{\dfrac{12\times 5+144}{65}} \\
 & \Rightarrow \dfrac{\dfrac{\left( 50+468 \right)}{65}}{\dfrac{\left( 60+144 \right)}{65}} \\
\end{align}\]
We can further write it as,
$\begin{align}
  & \dfrac{518\times 65}{204\times 65} \\
 & \Rightarrow \dfrac{518}{204} \\
 & \Rightarrow \dfrac{259}{102} \\
\end{align}$
Hence, we can say that for, $\sin \theta =\dfrac{12}{13}$, we get the value of $\dfrac{2\cos \theta +3\tan \theta }{\sin \theta +\tan \theta \sin \theta }$ as $\dfrac{259}{102}$.
Therefore, option (c) is the correct answer.

Note: While solving this question, the possible mistake one can make is a calculation mistake. Also, one can solve this question by using a few trigonometric properties like, $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$. By using these we can convert $\cos \theta $ and $\tan \theta $ to $\sin \theta $ and then we will put the values of $\sin \theta $ and then simplify to get the answer.