Question

Prove the following trigonometric relation: \[\dfrac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }=\dfrac{1}{3}\]

Answer 1

Hint:

In this question, from the given values of cot function by using the trigonometric identities we can find the values of the sine and cosine functions. Then on substituting the respective values in the given expression of the question we can calculate the left hand side value and the right hand side value. Then on comparing the values obtained, we get the result.

Complete step-by-step answer:

\[\begin{align}

  & \cot \theta =\dfrac{\cos \theta }{\sin \theta } \\

 & co{{\sec }^{2}}\theta -1={{\cot }^{2}}\theta \\

 & \cos ec\theta =\dfrac{1}{\sin \theta } \\

\end{align}\]


Now, from the given question we have

\[\begin{align}

  & 3\cot \theta =2 \\

 & \cot \theta =\dfrac{2}{3}\ \ \ \ \ ...(a) \\

\end{align}\]

Now, by using the trigonometric identity which gives the relation between the function that are mentioned in the hint, we get the following

\[\Rightarrow \cot \theta =\dfrac{\cos \theta }{\sin \theta }\]

Now, this can also be written as the following using the other relations given in the hint as follows

\[\begin{align}

  & \Rightarrow co{{\sec }^{2}}\theta -1={{\cot }^{2}}\theta \\

 & \Rightarrow {{\left( \dfrac{1}{\sin \theta } \right)}^{2}}-1={{\cot }^{2}}\theta \\

 & \Rightarrow {{\left( \dfrac{1}{\sin \theta } \right)}^{2}}={{\cot }^{2}}\theta +1 \\

\end{align}\]

Let us now substitute the value from the question and as well as from equation (a) in this

\[\begin{align}

  & \Rightarrow {{\left( \dfrac{1}{\sin \theta } \right)}^{2}}={{\left( \dfrac{2}{3} \right)}^{2}}+1 \\

 & \Rightarrow {{\left( \dfrac{1}{\sin \theta } \right)}^{2}}=\left( \dfrac{4}{9} \right)+1 \\

 & \Rightarrow {{\left( \dfrac{1}{\sin \theta } \right)}^{2}}=\dfrac{13}{9} \\

\end{align}\]

Now, this can be further written as

\[\begin{align}

  & \Rightarrow \left( \dfrac{1}{\sin \theta } \right)=\sqrt{\dfrac{13}{9}} \\

 & \Rightarrow \left( \dfrac{1}{\sin \theta } \right)=\sqrt{\dfrac{13}{9}} \\

 & \Rightarrow \sin \theta =\sqrt{\dfrac{9}{13}} \\

 & \Rightarrow \sin \theta =\dfrac{3}{\sqrt{13}} \\

\end{align}\]

Now, using the relation between the sin and cos function, we have

\[\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Now, this can be used to get the expression which can be written as

\[\begin{align}

  & \Rightarrow {{\left( \dfrac{3}{\sqrt{13}} \right)}^{2}}+{{\cos }^{2}}\theta =1 \\

 & \Rightarrow \dfrac{9}{13}+{{\cos }^{2}}\theta =1 \\

 & \Rightarrow {{\cos }^{2}}\theta =1-\dfrac{9}{13} \\

 & \Rightarrow {{\cos }^{2}}\theta =\dfrac{4}{13} \\

 & \Rightarrow \cos \theta =\dfrac{2}{\sqrt{13}} \\

\end{align}\]

Now, from the given expression in the question, on substituting the values, we have

\[\dfrac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }=\dfrac{1}{3}\]

Let us first consider the left hand side and calculate its value

\[\begin{align}

  & L.H.S=\dfrac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta } \\

 & L.H.S=\dfrac{4\times \dfrac{3}{\sqrt{13}}-3\times \dfrac{2}{\sqrt{13}}}{2\times \dfrac{3}{\sqrt{13}}+6\times \dfrac{2}{\sqrt{13}}} \\

 & L.H.S=\dfrac{\dfrac{12}{\sqrt{13}}-\dfrac{6}{\sqrt{13}}}{\dfrac{6}{\sqrt{13}}+\dfrac{12}{\sqrt{13}}} \\

 & L.H.S=\dfrac{\dfrac{6}{\sqrt{13}}}{\dfrac{18}{\sqrt{13}}} \\

 & L.H.S=\dfrac{1}{3} \\

\end{align}\]

Thus, the value of right hand side is equal to left hand side

Hence, it is verified that \[\dfrac{4\sin \theta -3\cos \theta }{2\sin \theta +6\cos \theta }=\dfrac{1}{3}\]


Note:
Instead of calculating the values of right hand side and left hand side by substituting the respective values we can calculate either of them and then use proper trigonometric identities that both the expressions are equal.
It is important to note that while calculating the values of respective functions we need to use the identities accordingly and solve them. Because neglecting any of the terms or writing it incorrectly changes the complete result.