Solveeit Logo
Login

Question

Question: Write the value of \({{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)\) ....

Write the value of tan⁡−13−sec⁡−1(−2){{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right) .

Explanation

Solution

The above question is related to inverse trigonometric function and for solving the problem, you just have to put the values of sec⁡−1(−2) and tan−13{{\sec }^{-1}}\left( -2 \right)\text{ and }ta{{n}^{-1}}\sqrt{3} and solve the expression by finding the difference between them. Remember for finding the value of sec⁡−1(−2){{\sec }^{-1}}\left( -2 \right) , you will have to use the property that sec⁡−1(−x)=π−sec⁡−1x{{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}x .

Complete step-by-step answer:
Before starting with the solution to the above question, we will first talk about the required details of different inverse trigonometric ratios. So, we must remember that inverse trigonometric ratios are completely different from trigonometric ratios and have many constraints related to their range and domain. So, to understand these constraints and the behaviour of inverse trigonometric functions, let us look at some of the important graphs. First, let us see the graph of sin⁡−1x{{\sin }^{-1}}x .

Now let us draw the graph of cos−1xco{{s}^{-1}}x .

Also, we will draw the graph of tan⁡−1x{{\tan }^{-1}}x as well.

So, looking at the above graphs, we can draw the conclusion that tan⁡−1x{{\tan }^{-1}}x is defined for all real values of x, i.e., the domain of the function tan⁡−1x{{\tan }^{-1}}x is all real numbers while its range comes out to be (−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right) . Unlike tan⁡−1x{{\tan }^{-1}}x the functions sin−1x and cos−1xsi{{n}^{-1}}x\text{ and co}{{\text{s}}^{-1}}x have the is defined only for x∈[−1,1]x\in [-1,1] .
Now moving to the solution to the above question, we will start with the simplification of the expression given in the question.
tan⁡−13−sec⁡−1(−2){{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)
We know that sec⁡−1(−x)=π−sec⁡−1x{{\sec }^{-1}}\left( -x \right)=\pi -{{\sec }^{-1}}x , and −2-2 also lies in the domain of sec⁡−1x{{\sec }^{-1}}x . So, using this value in our expression, we get
tan⁡−13−(π−sec⁡−12){{\tan }^{-1}}\sqrt{3}-\left( \pi -{{\sec }^{-1}}2 \right)
We also know that tan−13=π3ta{{n}^{-1}}\sqrt{3}=\dfrac{\pi }{3} and sec⁡−12=π3{{\sec }^{-1}}2=\dfrac{\pi }{3} , and 3\sqrt{3} also lies in the domain of tan−1xta{{n}^{-1}}x . So, using this value in our expression, we get
π3−(π−π3)=π3−2π3=−π3\dfrac{\pi }{3}-\left( \pi -\dfrac{\pi }{3} \right)=\dfrac{\pi }{3}-\dfrac{2\pi }{3}=-\dfrac{\pi }{3}
Therefore, the value of tan⁡−13−sec⁡−1(−2){{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right) is equal to −π3-\dfrac{\pi }{3} .

Note: Be careful about the range and domain of different trigonometric inverse functions as they are very confusing and may lead to errors. Don’t miss the final negative sign while reporting the answer. It is also important that you learn the trigonometric table which is as follow: