Question

Find the principal value of ${{\sec }^{-1}}\left( 2\tan \left( \dfrac{3\pi }{4} \right) \right)$

Answer 1

Hint: 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.