Find the principal value of \[{{\sec }^{-1}}\left( -\sqrt{2} \right)\].
VerifiedHint: First of all, use \[{{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}\left( x \right)\]. Now take sec on both sides and use \[\sec \left( \pi -\theta \right)=-\sec \theta \]. Now use the trigonometric table to find the angle at \[\sec \theta =\sqrt{2}\] and from this find the principal value of the given expression.
Complete step-by-step answer:
Here, we have to find the principal value of \[{{\sec }^{-1}}\left( -\sqrt{2} \right)\]. Let us take the principal value of \[{{\sec }^{-1}}\left( -\sqrt{2} \right)\] as y. So, we get,
\[y={{\sec }^{-1}}\left( -\sqrt{2} \right)\]
We know that \[{{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}x\]. By using this in the RHS of the above equation, we get,
\[y=\pi -{{\sec }^{-1}}\left( \sqrt{2} \right)\]
By taking sec on both sides of the above equation, we get,
\[\sec y=\sec \left( \pi -{{\sec }^{-1}}\left( \sqrt{2} \right) \right)\]
We know that \[\sec \left( \pi -\theta \right)=-\sec \theta \]. By using this in the RHS of the above equation, we get,
\[\sec y=-\sec \left( {{\sec }^{-1}}\left( \sqrt{2} \right) \right)\]
We know that \[\sec \left( {{\sec }^{-1}}x \right)=x\]. By using it in the RHS of the above equation, we get,
\[\sec y=-\sqrt{2}....\left( i \right)\]
From the above table, we can see that
$\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}$ and we know that $\cos\theta=\dfrac{1}{\sec\theta}$
So, \[\sec \dfrac{\pi }{4}=\sqrt{2}\]
By multiplying – 1 on both the sides, we get,
\[-\sec \dfrac{\pi }{4}=-\sqrt{2}\]
We know that \[\sec \left( \pi -\theta \right)=-\sec \theta \]. So, we can write \[-\sec \dfrac{\pi }{4}\] as \[\sec \left( \pi -\dfrac{\pi }{4} \right)\]. So, we get,
\[\sec \left( \pi -\dfrac{\pi }{4} \right)=-\sqrt{2}\]
\[\sec \left( \dfrac{3\pi }{4} \right)=-\sqrt{2}\]
By substituting the value of \[-\sqrt{2}\] in equation (i). We get,
\[\sec y=\sec \left( \dfrac{3\pi }{4} \right)\]
We know that the range of \[{{\sec }^{-1}}x\] for the principal values is \[\left[ 0,\pi \right]-\dfrac{\pi }{2}\]. So, we get, \[y=\dfrac{3\pi }{4}\].
Hence, we get the value of \[{{\sec }^{-1}}\left( -\sqrt{2} \right)\] as \[\dfrac{3\pi }{4}\].
Note: In this question, as we know that \[\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}\]. So, we get, \[{{\sec }^{-1}}\left( \sqrt{2} \right)=\dfrac{\pi }{4}\]. So, we can substitute the value of \[{{\sec }^{-1}}\sqrt{2}\] in \[y=\pi -{{\sec }^{-1}}\left( \sqrt{2} \right)\] initially to get \[y=\dfrac{3\pi }{4}\] without doing so many steps. Also, students must remember that y must lie in the range of \[{{\sec }^{-1}}x\] which is \[\left[ 0,\pi \right]-\dfrac{\pi }{2}\]. 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.
