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Question: Find the principal value of \[{{\sec }^{-1}}\left( -\sqrt{2} \right)\]....

Find the principal value of sec⁡−1(−2){{\sec }^{-1}}\left( -\sqrt{2} \right).

Explanation

Solution

Hint: First of all, use sec⁡−1(−x)=π−sec⁡−1(x){{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}\left( x \right). Now take sec on both sides and use sec⁡(π−θ)=−sec⁡θ\sec \left( \pi -\theta \right)=-\sec \theta . Now use the trigonometric table to find the angle at sec⁡θ=2\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(−2){{\sec }^{-1}}\left( -\sqrt{2} \right). Let us take the principal value of sec⁡−1(−2){{\sec }^{-1}}\left( -\sqrt{2} \right) as y. So, we get,

y=sec⁡−1(−2)y={{\sec }^{-1}}\left( -\sqrt{2} \right)

We know that sec⁡−1(−x)=π−sec⁡−1x{{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}x. By using this in the RHS of the above equation, we get,

y=π−sec⁡−1(2)y=\pi -{{\sec }^{-1}}\left( \sqrt{2} \right)

By taking sec on both sides of the above equation, we get,

sec⁡y=sec⁡(π−sec⁡−1(2))\sec y=\sec \left( \pi -{{\sec }^{-1}}\left( \sqrt{2} \right) \right)

We know that sec⁡(π−θ)=−sec⁡θ\sec \left( \pi -\theta \right)=-\sec \theta . By using this in the RHS of the above equation, we get,

sec⁡y=−sec⁡(sec⁡−1(2))\sec y=-\sec \left( {{\sec }^{-1}}\left( \sqrt{2} \right) \right)

We know that sec⁡(sec⁡−1x)=x\sec \left( {{\sec }^{-1}}x \right)=x. By using it in the RHS of the above equation, we get,

sec⁡y=−2....(i)\sec y=-\sqrt{2}....\left( i \right)

From the above table, we can see that

cos⁡π4=12\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}} and we know that cos⁡θ=1sec⁡θ\cos\theta=\dfrac{1}{\sec\theta}

So, sec⁡π4=2\sec \dfrac{\pi }{4}=\sqrt{2}

By multiplying – 1 on both the sides, we get,

−sec⁡π4=−2-\sec \dfrac{\pi }{4}=-\sqrt{2}

We know that sec⁡(π−θ)=−sec⁡θ\sec \left( \pi -\theta \right)=-\sec \theta . So, we can write −sec⁡π4-\sec \dfrac{\pi }{4} as sec⁡(π−π4)\sec \left( \pi -\dfrac{\pi }{4} \right). So, we get,

sec⁡(π−π4)=−2\sec \left( \pi -\dfrac{\pi }{4} \right)=-\sqrt{2}

sec⁡(3π4)=−2\sec \left( \dfrac{3\pi }{4} \right)=-\sqrt{2}

By substituting the value of −2-\sqrt{2} in equation (i). We get,

sec⁡y=sec⁡(3π4)\sec y=\sec \left( \dfrac{3\pi }{4} \right)

We know that the range of sec⁡−1x{{\sec }^{-1}}x for the principal values is [0,π]−π2\left[ 0,\pi \right]-\dfrac{\pi }{2}. So, we get, y=3π4y=\dfrac{3\pi }{4}.

Hence, we get the value of sec⁡−1(−2){{\sec }^{-1}}\left( -\sqrt{2} \right) as 3π4\dfrac{3\pi }{4}.

Note: In this question, as we know that sec⁡(π4)=2\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}. So, we get, sec⁡−1(2)=π4{{\sec }^{-1}}\left( \sqrt{2} \right)=\dfrac{\pi }{4}. So, we can substitute the value of sec⁡−12{{\sec }^{-1}}\sqrt{2} in y=π−sec⁡−1(2)y=\pi -{{\sec }^{-1}}\left( \sqrt{2} \right) initially to get y=3π4y=\dfrac{3\pi }{4} without doing so many steps. Also, students must remember that y must lie in the range of sec⁡−1x{{\sec }^{-1}}x which is [0,π]−π2\left[ 0,\pi \right]-\dfrac{\pi }{2}. Students must know that the values of trigonometric ratios like sin⁡θ,cos⁡θ\sin \theta ,\cos \theta , etc. at general angles like 0,30o,45o,60o,90o0,{{30}^{o}},{{45}^{o}},{{60}^{o}},{{90}^{o}} to easily solve the question.