Question

For the principal value, evaluate the following:
${{\sec }^{-1}}\left( \sqrt{2} \right)+2{{\operatorname{cosec}}^{-1}}\left( -\sqrt{2} \right)$.

Answer 1

Hint:For the above question we will have to know about the principal value of an inverse trigonometric function is a value that belongs to the principal branch of range of function. We know that the function branch of range for ${{\sec }^{-1}}x$ is \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\} and for ${{\operatorname{cosec}}^{-1}}x$ is \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}. We can start solving this question by taking $\theta ={{\sec }^{-1}}\left( \sqrt{2} \right)$ and then find the principal value of $\theta$. Then we can proceed to ${{\operatorname{cosec}}^{-1}}\left( -\sqrt{2} \right)$ in a similar way.

Complete step-by-step answer:
We have been given the trigonometric expression ${{\sec }^{-1}}\left( \sqrt{2} \right)+2{{\operatorname{cosec}}^{-1}}\left( -\sqrt{2} \right)$.
Now we know that the principal value means the value which lies between the defined range of inverse trigonometric functions.
For ${{\sec }^{-1}}x$ the range is \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}.
For ${{\operatorname{cosec}}^{-1}}x$ the range is \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}.
Let us suppose $\theta ={{\sec }^{-1}}\sqrt{2}$.
Taking secant function on both the sides of the expression, we get as follows:
$\sec \theta =\sec \left( {{\sec }^{-1}}\sqrt{2} \right)$
We know that $\sec \left( {{\sec }^{-1}}x \right)=x$, where x must lie between the interval $\left[ -\infty ,-1 \right]\cup \left[ 1,\infty \right]$.
$\Rightarrow \sec \theta =\sqrt{2}$
We know that $\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}$.
$\Rightarrow \theta =\dfrac{\pi }{4}$
Also, $\theta =\dfrac{\pi }{4}$ belongs to the interval \left[ 0,\pi \right]-\left\\{ \dfrac{\pi }{2} \right\\}.
Hence ${{\sec }^{-1}}\sqrt{2}=\dfrac{\pi }{4}$.
Again, let us suppose $\theta ={{\operatorname{cosec}}^{-1}}\left( -\sqrt{2} \right)$
Taking cosecant function on both the sides of the equation, we get as follows:
$\operatorname{cosec}\theta =\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}\left(-\sqrt{2} \right)\right)$
We know that $\operatorname{cosec}\left( {{\operatorname{cosec}}^{-1}}x \right)=x$, where x must lie between the interval $\left( -\infty ,-1 \right]\cup \left[ 1,\infty \right)$.
$\Rightarrow \operatorname{cosec}\theta =-\sqrt{2}$
We know that $\operatorname{cosec}\left( \dfrac{-\pi }{4} \right)=-\sqrt{2}$
$\Rightarrow \theta =\dfrac{-\pi }{4}$
Also, $\theta =\dfrac{-\pi }{4}$ belongs to the interval \left[ \dfrac{-\pi }{2},\dfrac{\pi }{2} \right]-\left\\{ 0 \right\\}.
Hence $cose{{c}^{-1}}\left( \sqrt{2} \right)=\dfrac{-\pi }{4}$.
On substituting the value of ${{\sec }^{-1}}\sqrt{2}=\dfrac{\pi }{4}$ and $cose{{c}^{-1}}\left( \sqrt{2} \right)=\dfrac{-\pi }{4}$ in the above equation, we get as follows:
$\Rightarrow {{\sec }^{-1}}\sqrt{2}+2cose{{c}^{-1}}\left( \sqrt{2} \right)=\dfrac{\pi }{4}+2\left( \dfrac{-\pi }{4} \right)=\dfrac{\pi }{4}-\dfrac{2\pi }{4}=\dfrac{\pi }{4}-\dfrac{\pi }{2}=\dfrac{\pi -2\pi }{4}=\dfrac{-\pi }{4}$
Therefore, the principal value of the given expression ${{\sec }^{-1}}\sqrt{2}+2cose{{c}^{-1}}\left( \sqrt{2} \right)$ is equal to $\left( \dfrac{-\pi }{4} \right)$.

Note: Be careful while finding the principal value of the inverse trigonometric function and do check once that the value must lie between the principal branch of range of the function. Sometimes we forget the ‘2’ multiplied by $cose{{c}^{-1}}\left( \sqrt{2} \right)$ in the given expression and we just substitute the values of $cose{{c}^{-1}}\left( \sqrt{2} \right)$ and we get the incorrect answer. So be careful while solving.