Question

If $3\cos \theta =5\sin \theta $, then the value of $\dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec }^{3}}\theta -2\cos \theta }$ is

a)$\dfrac{271}{979}$
b)$\dfrac{316}{2937}$
c)$\dfrac{542}{2937}$
d)None of these

Answer 1

Hint: Here, first we have to find the value of $\dfrac{\sin \theta }{\cos \theta }$ from $3\cos \theta =5\sin \theta $. Then, divide the numerator and denominator of the expression $\dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec }^{3}}\theta -2\cos \theta }$ by $\cos \theta $. Now, to find the value of ${{\sec }^{4}}\theta $ first we have to find the value of ${{\cos }^{4}}\theta $ from $3\cos \theta =5\sin \theta $.

Complete step-by-step answer:
Here, we are given that $3\cos \theta =5\sin \theta $.

Now, we have to find the value of $\dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec }^{3}}\theta -2\cos \theta }$.

First consider the expression

$3\cos \theta =5\sin \theta $

Next, by cross multiplication we get:

$\Rightarrow \dfrac{3}{5}=\dfrac{\sin \theta }{\cos \theta }$

Now, consider the expression $\dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec }^{3}}\theta -2\cos \theta }$, dividing the numerator and denominator by $\cos \theta $ we obtain,

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{5\dfrac{\sin \theta }{\cos \theta }-2\dfrac{{{\sec

}^{3}}\theta }{\cos \theta }+2\dfrac{\cos \theta }{\cos \theta }}{5\dfrac{\sin \theta }{\cos

 \theta }+2\dfrac{{{\sec }^{3}}\theta }{\cos \theta }-2\dfrac{\cos \theta }{\cos \theta }}\]

Next, by cancellation, we obtain,

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{5\dfrac{\sin \theta }{\cos \theta }-2\dfrac{{{\sec

}^{3}}\theta }{\cos \theta }+2}{5\dfrac{\sin \theta }{\cos \theta }+2\dfrac{{{\sec }^{3}}\theta

}{\cos \theta }-2}\]

Now, by substituting the value of $\dfrac{\sin \theta }{\cos \theta }=\dfrac{3}{5}$ in the

 above expression, we obtain,

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{5\times \dfrac{3}{5}-2\dfrac{{{\sec }^{3}}\theta }{\cos

\theta }+2}{5\times \dfrac{3}{5}+2\dfrac{{{\sec }^{3}}\theta }{\cos \theta }-2}\]

In the next step, again by cancellation, we obtain:

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{3-2\dfrac{{{\sec }^{3}}\theta }{\cos \theta

}+2}{3+2\dfrac{{{\sec }^{3}}\theta }{\cos \theta }-2}\]

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{5-2\dfrac{{{\sec }^{3}}\theta }{\cos \theta

}}{1+2\dfrac{{{\sec }^{3}}\theta }{\cos \theta }}\]

We know that $\sec \theta =\dfrac{1}{\cos \theta }$, by substituting this in the above

expression we get,

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{5-2{{\sec }^{3}}\theta \times \sec \theta }{1+2{{\sec

}^{3}}\theta \times \sec \theta }\]

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{5-2{{\sec }^{4}}\theta }{1+2{{\sec }^{4}}\theta }\]

                           ……. (1)
Now, consider

$3\cos \theta =5\sin \theta $

Then, by cross multiplication we get,

$\Rightarrow \cos \theta =\dfrac{5}{3}\sin \theta $

Next, by squaring we get,

$\begin{align}

  & \Rightarrow {{\cos }^{2}}\theta ={{\left( \dfrac{5}{3}\sin \theta \right)}^{2}} \\

 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{25}{9}{{\sin }^{2}}\theta \\

\end{align}$

We know that,

${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $

Now, by substituting this in the above expression we get,

$\Rightarrow {{\cos }^{2}}\theta =\dfrac{25}{9}\left( 1-{{\cos }^{2}}\theta \right)$

$\begin{align}

  & \Rightarrow {{\cos }^{2}}\theta =\dfrac{25}{9}-\dfrac{25}{9}\times {{\cos }^{2}}\theta \\

 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{25}{9}-\dfrac{25}{9}{{\cos }^{2}}\theta \\

\end{align}$

Now, take $\dfrac{25}{9}{{\cos }^{2}}\theta $ to the left side, we get,

$\Rightarrow {{\cos }^{2}}\theta +\dfrac{25}{9}{{\cos }^{2}}\theta =\dfrac{25}{9}$

Next, by taking the LCM we obtain,

$\begin{align}

  & \Rightarrow \dfrac{9{{\cos }^{2}}\theta +25{{\cos }^{2}}\theta }{9}=\dfrac{25}{9} \\

 & \Rightarrow \dfrac{34{{\cos }^{2}}\theta }{9}=\dfrac{25}{9} \\

\end{align}$

Next, by cross multiplication, we get,

$\Rightarrow {{\cos }^{2}}\theta =\dfrac{25}{9}\times \dfrac{9}{34}$

In the next step, by cancellation, we obtain,

$\Rightarrow {{\cos }^{2}}\theta =\dfrac{25}{34}$

Now, by squaring again we obtain,

$\Rightarrow {{\cos }^{4}}\theta =\dfrac{625}{1156}$

We know that,

$\begin{align}

  & {{\sec }^{4}}\theta =\dfrac{1}{{{\cos }^{4}}\theta } \\

 & \Rightarrow {{\sec }^{4}}\theta =\dfrac{1}{\dfrac{625}{1156}} \\

 & \Rightarrow {{\sec }^{4}}\theta =1\times \dfrac{1156}{625} \\

 & \Rightarrow {{\sec }^{4}}\theta =\dfrac{1156}{625} \\

\end{align}$

Now, by substituting the value of ${{\sec }^{4}}\theta =\dfrac{1156}{625}$ in equation (1) we obtain:

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{5-2\times \dfrac{1156}{625}}{1+2\times

\dfrac{1156}{625}}\]

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{5-\dfrac{2312}{625}}{1+\dfrac{2312}{625}}\]

Next, by taking the LCM we obtain:

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{\dfrac{3125-2312}{625}}{\dfrac{625+2312}{625}}\]

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{\dfrac{813}{625}}{\dfrac{2937}{625}}\]

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{813}{625}\times \dfrac{625}{2937}\]

Next, by cancellation, we get:

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{813}{2937}\]

In the next step by cancellation we get:

\[\Rightarrow \dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec

}^{3}}\theta -2\cos \theta }=\dfrac{271}{979}\]

Therefore, we can say that when $3\cos \theta =5\sin \theta $, then the value of

$\dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec }^{3}}\theta

-2\cos \theta }=\dfrac{271}{979}$

Hence, the correct answer for this question is option (a).


Note: Here, we can also solve this problem by writing $\tan \theta =\dfrac{\sin \theta }{\cos \theta }=\dfrac{3}{5}$. Hence, we have $\tan \theta =\dfrac{Opposite\text{ }side}{Adjacent\text{ }side}$, we have to find hypotenuse with the help of Pythagoras theorem, that is the square of hypotenuse is the sum of the squares of opposite side and adjacent side. Then with the help of hypotenuse, find the value of $\sin \theta $ and $\cos \theta $. From $\cos \theta $ we can find the value of $\sec \theta $ where $\sec \theta =\dfrac{1}{\cos \theta }$. After that find ${{\sec }^{4}}\theta $. Substitute all these values in the expression $\dfrac{5\sin \theta -2{{\sec }^{3}}\theta +2\cos \theta }{5\sin \theta +2{{\sec }^{3}}\theta -2\cos \theta }$ to get the required answer.