
How do I simplify $\sin \left( {\left( {\tan ^{- 1}} \right)x} \right)$?
Answer
558.3k+ views
Hint: In order to determine the simplification of the above question, let the ${\tan ^{ - 1}}x = \theta $, and recall that $\sin \theta $ can be written as $\dfrac{{\tan \theta }}{{\sec \theta }}$. Now replacing the $\sec \theta $ in the denominator using the identity of trigonometry which state that ${\sec ^2}\theta = {\tan ^2}\theta + 1$ and after putting this put $x = \tan \theta $ as we have assumed earlier.
Formula used:
$\sin \theta = \dfrac{{\tan \theta }}{{\sec \theta }}$
${\sec ^2}\theta = {\tan ^2}\theta + 1$
Complete step by step answer:
We are given a trigonometric expression $\sin ({\tan ^{ - 1}}(x))$
Let ${\tan ^{ - 1}}x = \theta $
Taking ${\tan ^{ - 1}}$ on the right-hand side we get
$x = \tan \theta $ ----------- (1)
As we know that the sine can be written as the ratio of the tangent and secant
$\therefore \sin \theta = \dfrac{{\tan \theta }}{{\sec \theta }}$-----------(2)
Since we know that identity of trigonometry which states that the square of secant is equal to the sum of square of tangent and one .i.e. ${\sec ^2}\theta = {\tan ^2}\theta + 1$
We can say $\sec \theta = \sqrt {{{\tan }^2}\theta + 1} $
Now putting $\sec \theta = \sqrt {{{\tan }^2}\theta + 1} $ in equation (2), we get
$ \Rightarrow \sin \theta = \dfrac{{\tan \theta }}{{\sqrt {{{\tan }^2}\theta + 1} }}$
Putting back the $\tan \theta = x$ in the RHS and $\theta = {\tan ^{ - 1}}x$ in LHS as we have assumed this in equation (1)
$ \Rightarrow \sin ({\tan ^{ - 1}}x) = \dfrac{x}{{\sqrt {{x^2} + 1} }}$
Therefore, the simplification of expression $\sin ({\tan ^{ - 1}}x)$ is equal to $\dfrac{x}{{\sqrt {{x^2} + 1} }}$
Note: In Mathematics the inverse trigonometric functions (every so often additionally called anti-trigonometric functions or cyclomatic function) are the reverse elements of the mathematical functions In particular, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are utilized to get a point from any of the point's mathematical proportions. Reverse trigonometric functions are generally utilized in designing, route, material science, and calculation.
Formula used:
$\sin \theta = \dfrac{{\tan \theta }}{{\sec \theta }}$
${\sec ^2}\theta = {\tan ^2}\theta + 1$
Complete step by step answer:
We are given a trigonometric expression $\sin ({\tan ^{ - 1}}(x))$
Let ${\tan ^{ - 1}}x = \theta $
Taking ${\tan ^{ - 1}}$ on the right-hand side we get
$x = \tan \theta $ ----------- (1)
As we know that the sine can be written as the ratio of the tangent and secant
$\therefore \sin \theta = \dfrac{{\tan \theta }}{{\sec \theta }}$-----------(2)
Since we know that identity of trigonometry which states that the square of secant is equal to the sum of square of tangent and one .i.e. ${\sec ^2}\theta = {\tan ^2}\theta + 1$
We can say $\sec \theta = \sqrt {{{\tan }^2}\theta + 1} $
Now putting $\sec \theta = \sqrt {{{\tan }^2}\theta + 1} $ in equation (2), we get
$ \Rightarrow \sin \theta = \dfrac{{\tan \theta }}{{\sqrt {{{\tan }^2}\theta + 1} }}$
Putting back the $\tan \theta = x$ in the RHS and $\theta = {\tan ^{ - 1}}x$ in LHS as we have assumed this in equation (1)
$ \Rightarrow \sin ({\tan ^{ - 1}}x) = \dfrac{x}{{\sqrt {{x^2} + 1} }}$
Therefore, the simplification of expression $\sin ({\tan ^{ - 1}}x)$ is equal to $\dfrac{x}{{\sqrt {{x^2} + 1} }}$
Note: In Mathematics the inverse trigonometric functions (every so often additionally called anti-trigonometric functions or cyclomatic function) are the reverse elements of the mathematical functions In particular, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are utilized to get a point from any of the point's mathematical proportions. Reverse trigonometric functions are generally utilized in designing, route, material science, and calculation.
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