Question

Explain inverse Trigonometric functions .

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Answer 1

Answer:

In mathematics, the inverse trigonometric functions are the inverse functions of the trigonometric functions. Specifically, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle's trigonometric ratios.

Answer 2

$\large\underline{\sf{Solution-}}$

We know that

For every bijective function f : A ⟼ B, there exist a bijection g : B ⟼ A defined by

If f( x ) = y then g ( y ) = x.

We also know that,

Al trigonometric functions are not bijective, so their inverse does not exist.

But if we restrict their domains and codomains, they can be made bijective and we are able to get their inverse.

Domain of the inverse Trigonometric functions :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Domain \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {sin}^{ - 1}x & \sf [ - 1,1] \\ \\ \sf {cos}^{ - 1}x & \sf [ - 1,1 ] \\ \\ \sf {tan}^{ - 1}x & \sf ( - \infty , \infty )\\ \\ \sf {cosec}^{ - 1}x & \sf ( - \infty , - 1] \: \cup \:[1, \infty ) \\ \\ \sf {sec}^{ - 1}x & \sf ( - \infty , - 1] \: \cup \:[1, \infty )\\ \\ \sf {cot}^{ - 1}x & \sf ( - \infty , \infty ) \end{array}} \\ \end{gathered}$

Range of Inverse Trigonometric functions :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {sin}^{ - 1}x & \sf \bigg[ - \dfrac{\pi}{2} ,\dfrac{\pi}{2} \bigg] \\ \\ \sf {cos}^{ - 1}x & \sf [ 0,\pi] \\ \\ \sf {tan}^{ - 1}x & \sf \bigg( - \dfrac{\pi}{2} , \dfrac{\pi}{2} \bigg)\\ \\ \sf {cosec}^{ - 1}x & \sf \bigg[ - \dfrac{\pi}{2} ,\dfrac{\pi}{2} \bigg] - \{0 \} \\ \\ \sf {sec}^{ - 1}x & \sf [0,\pi ] - \bigg \{\dfrac{\pi}{2} \bigg\} \\ \\ \sf {cot}^{ - 1}x & \sf ( 0 , \pi ) \end{array}} \\ \end{gathered}$

Properties of Inverse trigonometric functions :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf x \: \in \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {sin}^{ - 1}(sinx) = x & \sf \bigg[ - \dfrac{\pi}{2} ,\dfrac{\pi}{2} \bigg] \\ \\ \sf {cos}^{ - 1}(cosx) = x & \sf [ 0,\pi] \\ \\ \sf {tan}^{ - 1}(tanx) = x & \sf \bigg( - \dfrac{\pi}{2} , \dfrac{\pi}{2} \bigg)\\ \\ \sf {cosec}^{ - 1}(cosecx) = x & \sf \bigg[ - \dfrac{\pi}{2} ,\dfrac{\pi}{2} \bigg] - \{0 \} \\ \\ \sf {sec}^{ - 1}(secx) = x & \sf [0,\pi ] - \bigg \{\dfrac{\pi}{2} \bigg\} \\ \\ \sf {cot}^{ - 1}(cotx) = x & \sf ( 0 , \pi ) \end{array}} \\ \end{gathered}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf x \: \in \: \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sin({sin}^{ - 1}x) & \sf [ - 1,1] \\ \\ \sf cos({cos}^{ - 1}x) & \sf [ - 1,1 ] \\ \\ \sf tan({tan}^{ - 1}x) & \sf ( - \infty , \infty )\\ \\ \sf cosec({cosec}^{ - 1}x) & \sf ( - \infty , - 1] \: \cup \:[1, \infty ) \\ \\ \sf sec({sec}^{ - 1}x) & \sf ( - \infty , - 1] \: \cup \:[1, \infty )\\ \\ \sf cot({cot}^{ - 1}x) & \sf ( - \infty , \infty ) \end{array}} \\ \end{gathered}$

$\begin{gathered}\boxed{\begin{array}{c|c} \bf property & \bf x \: \in \: \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf {sin}^{ - 1}x + {cos}^{ - 1}x = \dfrac{\pi}{2} & \sf [ - 1, \: 1] \\ \\ \sf {tan}^{ - 1}x + {cot}^{ - 1}x = \dfrac{\pi}{2} & \sf ( - \infty , \infty ) \\ \\ \sf {sec}^{ - 1}x + {cosec}^{ - 1}x = \dfrac{\pi}{2} & \sf ( - \infty , - 1] \: \cup \:[1, \infty ) \end{array}} \\ \end{gathered}$