Question

If ${{\sin }^{-1}}x=\dfrac{\pi }{5}$, for some $x\in (-1,1)$, then the value of ${{\cos }^{-1}}x$ is ?
(a)$\dfrac{3\pi }{10}$.
(b) $\dfrac{5\pi }{10}$.
(c) $\dfrac{7\pi }{10}$.
(d) $\dfrac{9\pi }{10}$.

Answer 1

Hint: We have given the question which is based on inverse trigonometric functions. Inverse trigonometric functions are the inverse functions which are based on basic trigonometric functions. Inverse Trigonometric functions depend on the domain and range of the functions.To solve this question, first we know which equation can be used to solve the question, then we put the equation according to our question demands then by taking lcm and adding value, we find out the answer.

Complete step by step solution: 
We have given the question which is ${{\sin }^{-1}}x=\dfrac{\pi }{5}$ for some $x\in (-1,1)$,
We have to find out the value of ${{\cos }^{-1}}x$.
As the question is based on inverse trigonometric functions,
We know that there is inverse trigonometric equation which is defined below ,
${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$.
Now we put the value of ${{\sin }^{-1}}x$ (which is given in the question) in the above equation , we get
$\dfrac{\pi }{5}+{{\cos }^{-1}}x=\dfrac{\pi }{2}$.
Now we subtract $\dfrac{\pi }{5}$from both sides of the above equation , we get
$\dfrac{\pi }{5}+{{\cos }^{-1}}x-\dfrac{\pi }{5}=\dfrac{\pi }{2}-\dfrac{\pi }{5}$.
By taking the LCM in the above equation and solving the equation , we get
${{\cos }^{-1}}x=\dfrac{5\pi -2\pi }{10}$.
Now solving the above equation , we get
${{\cos }^{-1}}x=\dfrac{3\pi }{10}$.
Hence, the value of ${{\cos }^{-1}}x=\dfrac{3\pi }{10}$.

Note: In this question, students make the mistake in reminding the inverse trigonometric equations. There are some inverse trigonometric functions that are crucial not to remember but to have a deeper understanding of those properties. It is very important for a student to remember all the trigonometric equations or to know how to solve the equations so that the student makes no mistake in solving the question.