Question

Find the value of $\cos \left( {{\sec }^{-1}}x+\cos e{{c}^{-1}}x \right)$ , $\left| x \right|\ge 1$.

Answer 1

Hint: At first convert ${{\sec }^{-1}}\left( x \right)$ as ${{\cos }^{-1}}\left( \dfrac{1}{x} \right)$ and $\cos e{{c}^{-1}}\left( x \right)$ as ${{\sin }^{-1}}\left( \dfrac{1}{x} \right)$. And after that use the identity that ${{\sin }^{-1}}\left( \dfrac{1}{x} \right)+{{\cos }^{-1}}\left( \dfrac{1}{x} \right)$ is $\dfrac{\pi }{2}$ .Then reduce the given expression as $\cos \left( \dfrac{\pi }{2} \right)$ or 0.

Complete step-by-step answer:
In the question we are asked to find the value of $\cos \left( {{\sec }^{-1}}x+\cos e{{c}^{-1}}x \right)$ with a given condition $\left| x \right|\ge 1$ .

Before proceeding let us know what are inverse trigonometric functions. Inverse trigonometric functions are functions that are inverse functions of trigonometric functions. Specifically they are inverses of sine, cosine, tangent, cotangent, secant, cosecant functions, and are used to obtain an angle from any of the angles' trigonometric ratios.

These are certain notations which are used. Some of the most common notation is using arc $\sin \left( x \right)$ , arc $\cos \left( x \right)$ , arc $\tan \left( x \right)$ instead of ${{\sin }^{-1}}\left( x \right)$ , ${{\cos }^{-1}}\left( x \right)$ and ${{\tan }^{-1}}\left( x \right)$ . These arise from geometric relationships. When measuring in radius, an angle $\theta $ radius will correspond to an arc whose length is r $\theta $ , where r is radius of circle. Thus in the unit circle, “the arc whose cosine is x” is the same as “the angle whose cosine is x”, the length of the arc of circle in radii is the same as the measurement of angle in radius.

Now we are given an expression, $\cos \left( {{\sec }^{-1}}x+\cos e{{c}^{-1}}x \right)$ .

At first consider the expression and find the value of ${{\sec }^{-1}}x+\cos e{{c}^{-1}}x$ .

We know that ${{\sec }^{-1}}x={{\cos }^{-1}}\left( \dfrac{1}{x} \right)$ and $\cos e{{c}^{-1}}x={{\sin }^{-1}}\left( \dfrac{1}{x} \right)$ so, we can re-write the expression as –

${{\cos }^{-1}}\left( \dfrac{1}{x} \right)+{{\sin }^{-1}}\left( \dfrac{1}{x} \right)$ .

We know the fact that ${{\cos }^{-1}}\left( t \right)+{{\sin }^{-1}}\left( t \right)=\dfrac{\pi }{2}$ for all functions of t for which function ${{\cos }^{-1}}\left( t \right)+{{\sin }^{-1}}\left( t \right)$ is defined. Here $t=\dfrac{1}{x}$ and we are given that $\left| x \right|\ge 1$ . So, we can write the value of ${{\cos }^{-1}}\left( \dfrac{1}{x} \right)+{{\sin }^{-1}}\left( \dfrac{1}{x} \right)$ is \[\dfrac{\pi }{2}\] .

Hence the value of ${{\sec }^{-1}}x+\cos e{{c}^{-1}}x$ is \[\dfrac{\pi }{2}\].

So, the given expression reduces to $\cos \left( \dfrac{\pi }{2} \right)$ . Now by using the standard value table we can say that the value of $\cos \left( \dfrac{\pi }{2} \right)$ is 0.

Hence the value of the given expression $\cos \left( {{\sec }^{-1}}x+\cos e{{c}^{-1}}x \right)$
is 0.


Note: We know an identity that is ${{\sec }^{-1}}x+\cos e{{c}^{-1}}x=\dfrac{\pi }{2}$ . So, in the given expression we can directly substitute \[\dfrac{\pi }{2}\] instead of $\cos \left( {{\sec }^{-1}}x+\cos e{{c}^{-1}}x \right)$ to write it as $\cos \dfrac{\pi }{2}$ . After that we can directly say that value is 0 by the use of the standard value of the trigonometric table.