VerifiedHint: First of all, take sec on both the sides and use \[\sec \left( {{\sec }^{-1}}x \right)=x\]. Now write \[\dfrac{3\pi }{4}=\pi -\dfrac{\pi }{4}\] and use \[\tan \left( \pi -\theta \right)=-\tan \theta \] and from the table of trigonometric ratios, find the value of \[2\tan \dfrac{3\pi }{4}\]. Now take \[{{\sec }^{-1}}\] on both the sides and from the table find the value of \[{{\sec }^{-1}}\left( 2 \right)\] to find the principal value of the given expression.
Complete step-by-step answer:
Here, we have to find the principal value of \[{{\sec }^{-1}}\left( 2\tan \left( \dfrac{3\pi }{4} \right) \right)\]. Now let us consider the value of \[{{\sec }^{-1}}\left( 2\tan \left( \dfrac{3\pi }{4} \right) \right)\] as y. So, we get,
\[y={{\sec }^{-1}}\left( 2\tan \left( \dfrac{3\pi }{4} \right) \right)\]
By taking sec on both sides of the above equation, we get,
\[\sec y=\sec \left( {{\sec }^{-1}}\left( 2\tan \left( \dfrac{3\pi }{4} \right) \right) \right)\]
We know that \[\sec \left( {{\sec }^{-1}}\theta \right)=\theta \]. By using this in the RHS of the above equation, we get,
\[\sec y=\tan \left( \dfrac{3\pi }{4} \right)\]
We can also write the above equation as,
\[\sec y=2\tan \left( \pi -\dfrac{\pi }{4} \right)\]
We know that \[\tan \left( \pi -\theta \right)=-\tan \theta \]. By using this in the RHS of the above equation, we get,
\[\sec y=-2\tan \dfrac{\pi }{4}....\left( i \right)\]
Now, from the table of the general trigonometric ratios, let us find the value of \[\tan \left( \dfrac{\pi }{4} \right)\].
From the above table, we can see that \[\tan \dfrac{\pi }{4}=1\]. So, by substituting the value of \[\tan \dfrac{\pi }{4}\] in equation (i), we get,
\[\sec y=-2\left( 1 \right)\]
\[\sec y=-2\]
By taking \[{{\sec }^{-1}}\] on both the sides of the above equation, we get,
\[{{\sec }^{-1}}\left( \sec y \right)={{\sec }^{-1}}\left( -2 \right)\]
Since, \[{{\sec }^{-1}}\left( \sec y \right)=y\], we get,
\[y={{\sec }^{-1}}\left( -2 \right)\]
We know that \[{{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}x\]. By using this in the above equation, we get,
\[y=\pi -{{\sec }^{-1}}\left( 2 \right)\]
From the table, we know that
$\cos\dfrac{\pi}{3}=\dfrac{1}{2}$ and we know that $\cos\theta=\dfrac{1}{\sec\theta}$
We can write \[\sec \dfrac{\pi }{3}=2\]
So, we get, \[{{\sec }^{-1}}\left( 2 \right)=\dfrac{\pi }{3}\]. By using it in the above equation, we get,
\[y=\pi -\dfrac{\pi }{3}\]
\[y=\dfrac{2\pi }{3}\]
So, we get the principal value of \[y={{\sec }^{-1}}\left( 2\tan \left( \dfrac{3\pi }{4} \right) \right)\] as \[\dfrac{2\pi }{3}\].
Note: In this question, students can cross-check their answer by substituting \[y=\dfrac{2\pi }{3}\] and taking sec on both the sides of the initial equation and checking if LHS = RHS. Also, students must take care that range of \[{{\sec }^{-1}}\theta \] is \[\left[ 0,\pi \right]-\dfrac{\pi }{2}\]. So y must lie in this interval. Also, students must know that the values of trigonometric ratios like \[\sin \theta ,\cos \theta \], etc. at general angles like \[0,{{30}^{o}},{{45}^{o}},{{60}^{o}},{{90}^{o}}\] to easily solve the question.
