Question

Evaluate the expression ${{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)+\text{cose}{{\text{c}}^{-1}}\dfrac{2}{\sqrt{3}}.$

Answer 1

Hint: To answer this question, we need to know the standard angles for the basic trigonometric functions. We just substitute the values of the angles corresponding to the inverse of values for the given trigonometric functions. Then we add and subtract the given terms to obtain the answer.

Complete step by step solution:
In order to solve such questions, it is very important to know the standard angle values such as $0,\dfrac{\pi }{6},\dfrac{\pi }{4},\dfrac{\pi }{3},\dfrac{\pi }{2},\ldots \ldots $ of the basic trigonometric functions. We also need to know the inverse concept. We know that if $\sin \theta =a,$ then $\theta $ can be found by taking the inverse of sin function on the other side and it is given as,
$\Rightarrow \theta ={{\sin }^{-1}}a$
This is applicable to all the trigonometric functions. Each of these trigonometric functions have a set of defined values for these standard angles. Knowing them is very important to solve this question easily. We know that $\tan \dfrac{\pi }{3}=\sqrt{3},$ therefore, ${{\tan }^{-1}}\sqrt{3}=\dfrac{\pi }{3}.$ Similarly, for the second term, $\sec \left( -\dfrac{2\pi }{3} \right)=-2,$ and this lies in the second quadrant since it is negative. Therefore, ${{\sec }^{-1}}\left( -2 \right)=-\dfrac{2\pi }{3}.$ Similarly solving for the third term, $\text{cosec}\dfrac{\pi }{3}=\dfrac{2}{\sqrt{3}},$ therefore, $\text{cose}{{\text{c}}^{-1}}\left( \dfrac{2}{\sqrt{3}} \right)=\dfrac{\pi }{3}.$
The equation in the question is given as,
$\Rightarrow {{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)+\text{cose}{{\text{c}}^{-1}}\dfrac{2}{\sqrt{3}}$
Substituting the values obtained above for this above equation,
$\Rightarrow \dfrac{\pi }{3}-\dfrac{2\pi }{3}+\dfrac{\pi }{3}$
Adding the positive terms,
$\Rightarrow \dfrac{2\pi }{3}-\dfrac{2\pi }{3}$
Subtracting the above two terms,
$\Rightarrow 0$
Hence, the value of ${{\tan }^{-1}}\sqrt{3}-{{\sec }^{-1}}\left( -2 \right)+\text{cose}{{\text{c}}^{-1}}\dfrac{2}{\sqrt{3}}$ is 0.

Note: It is essential to know the basic trigonometric angles and their values in order to solve such questions. Not just for this question, but it is essential as the very basics for solving any trigonometric problem. It is important to note that the inverse of a trigonometric function is not the same as reciprocal of a trigonometric function. This means that ${{\sin }^{-1}}\theta $ is different from $\dfrac{1}{\sin \theta }=\text{cosec}\theta \text{.}$