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**NCERT mathematics class 12 chapter Inverse Trigonometric Functions Chapter Summary**. You can find the summary for the**chapter 2**of**NCERT class 12 mathematics**in this page. So is the case if you are looking for**NCERT class 12 Maths**related topic**Inverse Trigonometric Functions**. This page contains summary of the chapter. If you’re looking for exercise solutions, they’re available atInverse Trigonometric Functions – Summary

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The domains and ranges (principal value branches) of inverse trigonometric functions are given in the following table:

Functions

Domain

Range

(Principal Value Branches)

(Principal Value Branches)

{y = \sin^{-1} x}

[1, 1]

{\left[-\dfrac{π}{2}, \dfrac{π}{2}\right]}

{y = \cos^{-1} x}

[1, 1]

[0, π]

{y = \cosec^{-1} x}

R – (1, 1)

{\left[-\dfrac{π}{2}, \dfrac{π}{2}\right] - \{0\}}

{y = \sec^{-1} x}

R – (-1, 1)

{[0, π] - \left\{\dfrac{π}{2}\right\}}

{y = \tan^{-1} x}

R

{\left(-\dfrac{π}{2}, \dfrac{π}{2}\right)}

{y = \cot^{-1} x}

R

(0, π)

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Both {\sin^{-1} x} and {(\sin x)^{-1}} are different functions. It is important that we do not confuse {\sin^{-1} x} as {(\sin x)^{-1}}. In fact

{(\sin x)^{-1} = \dfrac{1}{\sin x}}

This is true for all other trigonometric functions.

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The value of an inverse trigonometric function which lies in its principal value branch is called the principal value of that inverse trigonometric functions.

For suitable values of domain we have

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{y = \sin^{-1} x}

⇒ {x = \sin y}

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{x = \sin y}

⇒ {y = \sin^{-1} x}

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{\sin (\sin^{-1} x) = x}

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{\sin^{-1} (\sin x) = x}

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{\sin^{-1} \left(\dfrac{1}{x}\right) = \cosec^{-1} x}

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{\cos^{-1} \left(\dfrac{1}{x}\right) = \sec^{-1} x}

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{\tan^{-1} \left(\dfrac{1}{x}\right) = \cot^{-1} x}

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{\cos^{-1} (-x) = π - \cos^{-1} x}

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{\sec^{-1} (-x) = π - \sec^{-1} x}

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{\cot^{-1} (-x) = π - \cot^{-1} x}

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{\sin^{-1} (-x) = -\sin^{-1} x}

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{\cosec^{-1} (-x) = -\cosec^{-1} x}

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{\tan^{-1} (-x) = -\tan^{-1} x}

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{\sin^{-1} x + \cos^{-1} x = \dfrac{π}{2}}

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{\cosec^{-1} x + \sec^{-1} x = \dfrac{π}{2}}

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{\tan^{-1} x + \cot^{-1} x = \dfrac{π}{2}}

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{\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\dfrac{x + y}{1 - xy}\right)}, {xy \lt 1}

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{\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left(\dfrac{x + y}{1 - xy}\right)}, {xy \gt 1}; {x, y \gt 0}

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{\tan^{-1} x - \tan^{-1} y = \tan^{-1} \left(\dfrac{x - y}{1 + xy}\right)}, {xy \gt -1}

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{2\tan^{-1} x}

=

{\sin^{-1} \dfrac{2x}{1 + x^2}}

=

{\cos^{-1} \dfrac{1 - x^2}{1 + x^2}}

=

{\tan^{-1} \dfrac{2x}{1 - x^2}}