Question

Solve the following Trigonometric function :
$ \left( m+2 \right)\sin \theta +\left( 2m-1 \right)\cos \theta =2m+1 $ if
\[\begin{align}
  & \left( A \right)\tan \theta =\dfrac{3}{4} \\
 & \left( B \right)\tan \theta =\dfrac{4}{3} \\
 & \left( C \right)\tan \theta =\dfrac{2m}{{{m}^{2}}-1} \\
 & \left( D \right)\tan \theta =\dfrac{2m}{{{m}^{2}}+1} \\
\end{align}\]

Answer 1

Hint:
 In this question, we are given a trigonometric equation and we need to find the value of $ \tan \theta $ . For this we will first change $ \sin \theta \text{ and }\sec \theta $ into $ \tan \theta $ and then form a quadratic equation in terms of $ \tan \theta $ . Solving it by splitting the middle term, we will get the value of .... We will use following properties:
 $ \begin{align}
  & \left( i \right)\dfrac{\sin \theta }{\cos \theta }=\tan \theta \\
 & \left( ii \right)\dfrac{1}{\cos \theta }=\sec \theta \\
 & \left( iii \right)1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \\
 & \left( iv \right){{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
 & \left( v \right){{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
\end{align} $

Complete step by step answer:
Here we are given the equation as, $ \left( m+2 \right)\sin \theta +\left( 2m-1 \right)\cos \theta =2m+1 $ .
From options, we see that we need to find the values of $ \tan \theta $. So let us solve the above equation to get the value of $ \tan \theta $.
Dividing the equation by $ \cos \theta $ on both sides we have,
 $ \dfrac{\left( m+2 \right)\sin \theta }{\cos \theta }+\dfrac{\left( 2m-1 \right)\cos \theta }{\cos \theta }=\dfrac{2m+1}{\cos \theta } $ .
We know that $ \dfrac{\sin \theta }{\cos \theta }=\tan \theta \text{ and }\dfrac{1}{\cos \theta }=\sec \theta $ so we get,
 $ \left( m+2 \right)\tan \theta +\left( 2m-1 \right)=\left( 2m+1 \right)\sec \theta $ .
Now let us square it on both sides we get,
 $ {{\left[ \left( m+2 \right)\tan \theta +\left( 2m-1 \right) \right]}^{2}}={{\left[ \left( 2m+1 \right)\sec \theta \right]}^{2}} $ .
Using $ {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab $ on the left side of the equation we get,
 $ {{\left( m+2 \right)}^{2}}{{\tan }^{2}}\theta +{{\left( 2m-1 \right)}^{2}}+2\left( 2m-1 \right)\left( m+2 \right)\tan \theta ={{\left( 2m+1 \right)}^{2}}{{\sec }^{2}}\theta $ .
We know that $ 1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ so using it we get,
 $ \begin{align}
  & {{\left( m+2 \right)}^{2}}{{\tan }^{2}}\theta +{{\left( 2m-1 \right)}^{2}}+2\left( 2m-1 \right)\left( m+2 \right)\tan \theta ={{\left( 2m+1 \right)}^{2}}\left( 1+{{\tan }^{2}}\theta \right) \\
 & \Rightarrow {{\left( m+2 \right)}^{2}}{{\tan }^{2}}\theta +{{\left( 2m-1 \right)}^{2}}+2\left( 2m-1 \right)\left( m+2 \right)\tan \theta ={{\left( 2m+1 \right)}^{2}}+{{\left( 2m+1 \right)}^{2}}{{\tan }^{2}}\theta \\
\end{align} $ .
Bringing $ {{\tan }^{2}}\theta $ term common and constant together we get,
 $ \begin{align}
  & \Rightarrow {{\left( m+2 \right)}^{2}}{{\tan }^{2}}\theta -{{\left( 2m+1 \right)}^{2}}{{\tan }^{2}}\theta +2\left( 2m-1 \right)\left( m+2 \right)\tan \theta +{{\left( 2m-1 \right)}^{2}}-{{\left( 2m+1 \right)}^{2}}=0 \\
 & \Rightarrow \left[ {{\left( m+2 \right)}^{2}}-{{\left( 2m+1 \right)}^{2}} \right]{{\tan }^{2}}\theta +\left[ 2\left( 2m-1 \right)\left( m+2 \right) \right]\tan \theta +\left[ {{\left( 2m-1 \right)}^{2}}-{{\left( 2m+1 \right)}^{2}} \right]=0 \\
\end{align} $ .
Now let us simplify our terms using $ {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\text{ and }{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab $ we get,
 $ \Rightarrow \left[ {{m}^{2}}+4+4m-4{{m}^{2}}-1-4m \right]{{\tan }^{2}}\theta +\left[ 4{{m}^{2}}-2m+8m-4 \right]\tan \theta +\left[ 4{{m}^{2}}+1-4m-4{{m}^{2}}-1-4m \right]=0 $
Simplifying and cancelling the terms we get,
\[\begin{align}
  & \left( 3-3{{m}^{2}} \right){{\tan }^{2}}\theta +\left( 4{{m}^{2}}+6m-4 \right)\tan \theta -8m=0 \\
 & \Rightarrow 3\left( 1-{{m}^{2}} \right){{\tan }^{2}}\theta +\left( 4{{m}^{2}}+6m-4 \right)\tan \theta -8m=0 \\
\end{align}\].
Now let us factorize it to find the values of $ \tan \theta $ .
Splitting the middle term we get,
\[3\left( 1-{{m}^{2}} \right){{\tan }^{2}}\theta +6m\tan \theta +\left( 4{{m}^{2}}-4 \right)\tan \theta -8m=0\].
Taking $ 3\tan \theta $ common from the first two terms and taking -4 common from the third and the fourth term we get,
\[3\tan \theta \left[ \left( 1-{{m}^{2}} \right)\tan \theta +2 \right]-4\left[ \left( 1-{{m}^{2}} \right)\tan \theta +2 \right]=0\].
Now taking \[\left[ \left( 1-{{m}^{2}} \right)\tan \theta +2 \right]\] common from both terms we get,
\[\left[ \left( 1-{{m}^{2}} \right)\tan \theta +2 \right]\left( 3\tan \theta -4 \right)=0\].
We know that both these factors should be equal to zero. Therefore,
\[\left( 1-{{m}^{2}} \right)\tan \theta +2\text{ and }3\tan \theta -4=0\].
Solving them one by one we get,
\[\left( 1-{{m}^{2}} \right)\tan \theta +2=0\].
Taking 2 to the other side,
\[\left( 1-{{m}^{2}} \right)\tan \theta =-2\].
Dividing both sides by \[\left( 1-{{m}^{2}} \right)\] we get,
\[\tan \theta =\dfrac{-2}{\left( 1-{{m}^{2}} \right)}\].
Taking negative sign common from the denominator we get,
\[\tan \theta =\dfrac{-2}{-\left( {{m}^{2}}-1 \right)}\Rightarrow \dfrac{2}{{{m}^{2}}-1}\].
Now putting \[3\tan \theta -4=0\Rightarrow 3\tan \theta =4\] dividing both sides by 3 we get,
\[\tan \theta =\dfrac{4}{3}\].
Hence we have obtained two values of \[\tan \theta \] which are,
\[\tan \theta =\dfrac{-2}{\left( 1-{{m}^{2}} \right)}\text{ and }\tan \theta =\dfrac{4}{3}\].
Hence option B and option C are the correct answer.

Note:
Students should note that a quadratic equation always gives two solutions. So we have obtained two values of \[\tan \theta \]. Take care of the signs while solving the equation. Make sure to take the whole square on the left side while taking the square on both sides in the second step.