Question

If $2{\tan ^{ - 1}}x = {\sin ^{ - 1}}k$ , What will be the value of $k?$

Answer 1

Hint: We can see that we have a trigonometric expression in the above question. So we will apply the trigonometric formula to solve this question. The given expression has an inverse function, so we will directly apply the formula of $2{\tan ^{ - 1}}x$ . We will substitute the value and then by comparing we can solve this question.

Formula used:
$2{\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$

Complete step-by-step answer:
Let us first understand the inverse trigonometric functions. The inverse trigonometric functions are also known as the anti-trigonometric functions or sometimes it is called the arcus functions or cyclometric functions. We write the inverse trigonometric functions with the power of $ - 1$ as the exponential power on the function.
We can also denote the inverse with writing the word $arc$in front of the ratio, such as we can write inverse sine as $\arcsin = {\sin ^{ - 1}}$ , similarly for cosine, the inverse function ${\cos ^{ - 1}}$ can be also be written as $\arccos $ .
Here we have been given in the question: $2{\tan ^{ - 1}}x = {\sin ^{ - 1}}k$
We can write the expression also as:
 ${\sin ^{ - 1}}k = 2{\tan ^{ - 1}}x$
Now we will substitute the value
 $2{\tan ^{ - 1}}x = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$ in the above equation, i.e.
${\sin ^{ - 1}}k = {\sin ^{ - 1}}\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$
We can see that we have ${\sin ^{ - 1}}$ in both the left-hand side and right-hand side of the equation, so it will get cancel out or by comparing both the sides of the equation, we have:
 $k = \left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$
Hence the required value of $k$ is $\left( {\dfrac{{2x}}{{1 + {x^2}}}} \right)$

Note: We should know the various inverse trigonometric formulas of sine, cosine, tangent, cotangent and other trigonometric ratios. We should note that the formula of $2{\tan ^{ - 1}}x$ is also in cosine function i.e.
 $2{\tan ^{ - 1}}x = {\cos ^{ - 1}}\left( {\dfrac{{1 - {x^2}}}{{1 + {x^2}}}} \right)$ .
We should understand the requirements of the question and then carefully apply the formula which is suitable for the solution.