Question

Find the value of \[\theta \] from the expression, \[\sec 4\theta -\sec 2\theta =2\]

Answer 1

Hint: First of all, convert the given expression into the cosine ratio by using the identity \[\sec \theta =\dfrac{1}{\cos \theta }\] . Now, use the formula \[2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\] and simplify it further. Similarly, use the identity \[\cos \left( -\theta \right)=\cos \theta \] and \[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] . At last, use \[\cos \dfrac{\pi }{2}=0\] , \[\cos \dfrac{3\pi }{2}=0\] , \[\cos \dfrac{5\pi }{2}=0\] , and find the general value of \[\theta \] .

Complete step-by-step answer:
According to the question, we are given an expression in terms of trigonometric ratio and we are asked to find the value of \[\theta \] .
The given expression is \[\sec 4\theta -\sec 2\theta =2\] ………………………………………………(1)
We can observe that the above equation requires more.
We know the identity that secant ratio is the reciprocal of cosine ratio, \[\sec \theta =\dfrac{1}{\cos \theta }\] ………………………………………..(2)
Now, from equation (1) and equation (2), we get
\[\begin{align}
  & \Rightarrow \dfrac{1}{\cos 4\theta }-\dfrac{1}{\cos 2\theta }=2 \\
 & \Rightarrow \dfrac{\cos 2\theta -\cos 4\theta }{\cos 2\theta \cos 4\theta }=2 \\
\end{align}\]
\[\Rightarrow \cos 2\theta -\cos 4\theta =2\cos 2\theta \cos 4\theta \] ……………………………………(3)
We also know the formula that \[2\cos A\cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\] ……………………………………(4)
 Now, using the formula shown in equation (4) and on simplifying equation (3), we get
\[\Rightarrow \cos 2\theta -\cos 4\theta =\cos \left( 2\theta +4\theta \right)+\cos \left( 2\theta -4\theta \right)\]
\[\Rightarrow \cos 2\theta -\cos 4\theta =\cos 6\theta +\cos \left( -2\theta \right)\] …………………………………………(5)
We also know the identity that \[\cos \left( -\theta \right)=\cos \theta \] ………………………………………………….(6)
From equation (5) and equation (6), we get
\[\begin{align}
  & \Rightarrow \cos 2\theta -\cos 4\theta =\cos 6\theta +\cos 2\theta \\
 & \Rightarrow -\cos 4\theta =\cos 6\theta \\
 & \Rightarrow -\cos 4\theta -\cos 6\theta =0 \\
\end{align}\]
\[\Rightarrow \cos 4\theta +\cos 6\theta =0\] ………………………………………….(7)
We also know the identity that \[\cos A+\cos B=2\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)\] ……………………………………..(8)
Using equation (8) and on simplifying equation (7), we get
\[\Rightarrow 2\cos \left( \dfrac{6\theta +4\theta }{2} \right)\cos \left( \dfrac{6\theta -4\theta }{2} \right)=0\]
\[\Rightarrow 2\cos 5\theta \cos \theta =0\]
So, either \[\cos 5\theta =0\] or \[\cos \theta =0\] …………………………………..(9)
Let us take \[\cos 5\theta =0\] …………………………………..(10)
We know that \[\cos \dfrac{\pi }{2}=0\] ………………………………………..(11)
Now, from equation (10) and equation (11), we get
\[\begin{align}
  & \Rightarrow \cos 5\theta =\cos \dfrac{\pi }{2} \\
 & \Rightarrow 5\theta ={{\cos }^{-1}}\left( \cos \dfrac{\pi }{2} \right) \\
 & \Rightarrow 5\theta =\dfrac{\pi }{2} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{\pi }{10}\] ………………………………………………(12)
We also know that \[\cos \dfrac{3\pi }{2}=0\] ………………………………………..(13)
From equation (10) and equation (13), we get
\[\begin{align}
  & \Rightarrow \cos 5\theta =\cos \dfrac{3\pi }{2} \\
 & \Rightarrow 5\theta ={{\cos }^{-1}}\left( \cos \dfrac{3\pi }{2} \right) \\
 & \Rightarrow 5\theta =\dfrac{3\pi }{2} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{3\pi }{10}\] ………………………………………………(14)
We also know that \[\cos \dfrac{5\pi }{2}=0\] ………………………………………..(15)
From equation (10) and equation (15), we get
\[\begin{align}
  & \Rightarrow \cos 5\theta =\cos \dfrac{5\pi }{2} \\
 & \Rightarrow 5\theta ={{\cos }^{-1}}\left( \cos \dfrac{5\pi }{2} \right) \\
 & \Rightarrow 5\theta =\dfrac{5\pi }{2} \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{5\pi }{10}\] ………………………………………………(16)
From equation (12), equation (14), and equation (16), we can get the general value of \[\theta \] i.e.,
\[\Rightarrow \theta =\dfrac{\left( 2n+1 \right)\pi }{10}\] where n = 0, 1, 2,………….. …………………………….(17)
Now, let us proceed with \[\cos \theta =0\] …………………………………..(18)
Similarly, from equation (11) and equation (18), we get
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \dfrac{\pi }{2} \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \cos \dfrac{\pi }{2} \right) \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{\pi }{2}\] ………………………………………………(19)
From equation (13) and equation (18), we get
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \dfrac{3\pi }{2} \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \cos \dfrac{3\pi }{2} \right) \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{3\pi }{10}\] ………………………………………………(20)
From equation (15) and equation (18), we get
\[\begin{align}
  & \Rightarrow \cos \theta =\cos \dfrac{5\pi }{2} \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \cos \dfrac{5\pi }{2} \right) \\
\end{align}\]
\[\Rightarrow \theta =\dfrac{5\pi }{2}\] ………………………………………………(21)
From equation (19), equation (20), and equation (21), we can get the general value of \[\theta \] i.e.,
\[\Rightarrow \theta =\dfrac{\left( 2n+1 \right)\pi }{2}\] where n = 0, 1, 2,………….. …………………………….(22)
From equation (17) and equation (22), we have the value of \[\theta \] .
Therefore, the value of \[\theta \] is \[\dfrac{\left( 2n+1 \right)\pi }{10}\] or \[\dfrac{\left( 2n+1 \right)\pi }{2}\] .

Note: Whenever this type of question appears where we have an expression in terms of trigonometric ratio. The best way to approach them is to convert them into the sine ratio or cosine ratio using appropriate identity conversions. This will make the calculations easy and reduce mistakes.