Question

Find the value of ${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}$ if ${{\sec }^{-1}}x+{{\csc }^{-1}}x=\dfrac{\pi }{2}$ .

Answer 1

Hint:To find the value of ${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}$ , we will use the property ${{\cos }^{-1}}\dfrac{1}{x}={{\sec }^{-1}}x$ and substitute the value to the previous equation. Using the given \[{{\sec }^{-1}}x={{\csc }^{-1}}y\] and the property ${{\sec }^{-1}}x+{{\csc }^{-1}}x=\dfrac{\pi }{2}$ the required value can be obtained.

Complete step by step answer:
We need to find the value of ${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}$ given that \[{{\sec }^{-1}}x={{\csc }^{-1}}y\] .
Let us consider the equation ${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}$ .
We know that ${{\cos }^{-1}}\dfrac{1}{x}={{\sec }^{-1}}x$ . Thus the above equation can be written as
${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}={{\sec }^{-1}}x+{{\sec }^{-1}}y$
It is given that \[{{\sec }^{-1}}x={{\csc }^{-1}}y\] . Hence, the above equation can be written as
${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}={{\sec }^{-1}}x+{{\csc }^{-1}}x$
We know from the properties of inverse trigonometric functions that ${{\sec }^{-1}}x+{{\csc }^{-1}}x=\dfrac{\pi }{2}$ .
Hence, we will get
${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}=\dfrac{\pi }{2}$ .

Note:
Secant and cosecant are inversely related. There can be a possibility of making errors in the properties ${{\cos }^{-1}}\dfrac{1}{x}={{\sec }^{-1}}x$ and ${{\sec }^{-1}}x+{{\csc }^{-1}}x=\dfrac{\pi }{2}$ . You have to be very thorough with all the trigonometric properties. The above question can also be solved in an alternate method. This is illustrated below:
Let us first consider the ranges of the functions \[{{\sec }^{-1}}x\]and \[{{\csc }^{-1}}y\] .
We know that the range of \[{{\sec }^{-1}}x=\left[ 0,\dfrac{\pi }{2} \right)\cup \left( \dfrac{\pi }{2},\pi \right]\] and the range of \[{{\csc }^{-1}}y=\left[ \dfrac{-\pi }{2},0 \right)\cup \left( 0,\dfrac{\pi }{2} \right]\]
The common of these is $\dfrac{\pi }{4}$ .
Thus we can write $x$ from ${{\sec }^{-1}}x$as shown below.
$x=\sec \left( \dfrac{\pi }{4} \right)=\sqrt{2}$
Similarly, from \[\cos e{{c}^{-1}}y\] we can write as follows.
\[y=\csc \left( \dfrac{\pi }{4} \right)=\sqrt{2}\]
Hence we can find the value of${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}$ .
First let us find the value of ${{\cos }^{-1}}\dfrac{1}{x}$
${{\cos }^{-1}}\dfrac{1}{x}={{\cos }^{-1}}\dfrac{1}{\sqrt{2}}=\dfrac{\pi }{4}$
Similarly, we can find the value of${{\cos }^{-1}}\dfrac{1}{y}$ .
${{\cos }^{-1}}\dfrac{1}{y}={{\cos }^{-1}}\dfrac{1}{\sqrt{2}}=\dfrac{\pi }{4}$
Substituting these values in ${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}$ , we will get
${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}=\dfrac{\pi }{4}+\dfrac{\pi }{4}=\dfrac{\pi }{2}$
Now let us add these. So we will get
${{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}=\dfrac{\pi }{2}$
Hence, the value of \[{{\cos }^{-1}}\dfrac{1}{x}+{{\cos }^{-1}}\dfrac{1}{y}=\dfrac{\pi }{2}\]