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Properties of Inverse Trig Functions Class 12

October 15, 2026 12th Mathematics(English), JEE MAINS Maths(English) No Comments
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1 1.Introduction to Properties of Inverse Trig Functions Class 12
1.1 2.Properties of Inverse Trig Functions
1.2 3.Practice Problems of Properties of Inverse Trig Functions for Students
1.2.1 4.Frequently Asked Questions Related to Properties of Inverse Trig Functions
1.2.2 Q:1.The inverse trigonometric function is not valid for which values?
1.2.3 Q:2.What is meant by inverse trigonometric function?
1.2.4 Q:3.Define mathematics according to FELIX KLEIN
1.2.5 5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

1.Introduction to Properties of Inverse Trig Functions Class 12

Learn about Properties of Inverse Trig Functions.Master its proofs,Main properties step-by-step and complete proofs.

Also Read This Article:- Principal Value: Inverse Trig Class 12

2.Properties of Inverse Trig Functions

(1.) \sin(\sin^{-1} x) = x, -1 \leq x \leq 1 \cdots(1)  
Let \sin^{-1} x = \theta then \sin \theta = x [by def]
Put \sin^{-1} x = \theta in (1), we get
\sin(\sin^{-1} x) = x
Now \theta = \sin^{-1} x, \ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}
or \theta = \sin^{-1}(\sin \theta)
Similarly according to given table:
\cos(\cos^{-1} x) = x \quad \cos^{-1}(\cos \theta) = \theta
\tan(\tan^{-1} x) = x \quad \tan^{-1}(\tan \theta) = \theta
\cot(\cot^{-1} x) = x \quad \cot^{-1}(\cot \theta) = \theta
\sec(\sec^{-1} x) = x \quad \sec^{-1}(\sec \theta) = \theta
\operatorname{cosec}(\operatorname{cosec}^{-1} x) = x \quad \operatorname{cosec}^{-1}(\operatorname{cosec} \theta) = \theta
(2.) \sin^{-1} \frac{1}{x} = \operatorname{cosec}^{-1} x, \ x \in \mathbb{R} \sim (-1, 1)
Proof: Let \sin^{-1}\left(\frac{1}{x}\right) = \theta \Rightarrow \sin \theta = \frac{1}{x} \cdots(1)
\Rightarrow \operatorname{cosec} \theta = x
\Rightarrow \theta = \operatorname{cosec}^{-1} x \cdots(2)
Put value of \theta in (2), we get
\sin^{-1}\left(\frac{1}{x}\right) = \operatorname{cosec}^{-1} x
Similarly
\operatorname{cosec}^{-1} x = \sin^{-1} \frac{1}{x}, \ x \leq -1, x \geq 1
\cos^{-1} x = \sec^{-1}\left(\frac{1}{x}\right), \ -1 \leq x \leq 1
\sec^{-1} x = \cos^{-1}\left(\frac{1}{x}\right), \ x \leq -1, x \geq 1
\tan^{-1} x = \cot^{-1} \frac{1}{x} \text{ and } \tan^{-1}\left(\frac{1}{x}\right), x > 0
(3.) \sin^{-1}(-x) = -\sin^{-1} x
and \cos^{-1}(-x) = \pi - \cos^{-1} x, -1 \leq x \leq 1
Proof: \sin^{-1}(-x) = \theta
\Rightarrow -x = \sin\theta
\Rightarrow x = -\sin\theta = \sin(-\theta)
\Rightarrow \sin^{-1} x = -\theta
\Rightarrow \sin^{-1} x = -\sin^{-1}(-x)
Similarly \cos^{-1}(-x) = \theta then x = -\cos\theta
\Rightarrow x = \cos(\pi - \theta)
\therefore \cos^{-1} x = \pi - \theta
\Rightarrow \cos^{-1}(x) = \pi - \cos^{-1}(-x)
\Rightarrow \cos^{-1}(-x) = \pi - \cos^{-1}(x)
Similarly
\tan^{-1}(-x) = -\tan^{-1} x, \quad \operatorname{cosec}^{-1}(-x) = -\operatorname{cosec}^{-1} x
\sec^{-1}(-x) = \pi - \sec^{-1} x, \quad \cot^{-1}(-x) = \pi - \cot^{-1} x

(4.) Prove that
(a) \sin^{-1} x \pm \sin^{-1} y = \sin^{-1} \leqft[ x\sqrt{1-y^2} \pm y\sqrt{1-x^2} \right]
Proof: Let \sin^{-1} x = \theta_1 \text{ i.e. } \sin\theta_1 = x
and \sin^{-1} y = \theta_2 \text{ i.e. } \sin\theta_2 = y
then \cos\theta_1 = \sqrt{1 - \sin^2\theta_1} = \sqrt{1 - x^2}
Similarly \cos\theta_2 = \sqrt{1 - \sin^2\theta_2} = \sqrt{1 - y^2}
We know that
\sin(\theta_1 \pm \theta_2) = \sin \theta_1 \cos\theta_2 \pm \cos\theta_1 \sin\theta_2
\Rightarrow \theta_1 \pm \theta_2 = \sin^{-1}(\sin\theta_1 \cos\theta_2 \pm \cos\theta_1 \sin\theta_2)
= \sin^{-1} \leqft[ x\sqrt{1-y^2} \pm y\sqrt{1-x^2} \right]
\Rightarrow \sin^{-1} x \pm \sin^{-1} y = \sin^{-1} \leqft[ x\sqrt{1-y^2} \pm y\sqrt{1-x^2} \right]
(b) 2\sin^{-1} x = \sin^{-1} [2x\sqrt{1-x^2}]
Proof: Let \sin^{-1} x = \theta \text{ i.e. } \sin\theta = x
\therefore \sin 2\theta = 2\sin\theta\cos\theta
= 2\sin\theta\sqrt{1-\sin^2\theta}
= 2x\sqrt{1-x^2}
\Rightarrow 2\theta = \sin^{-1}[2x\sqrt{1-x^2}]
\Rightarrow 2\sin^{-1} x = \sin^{-1}[2x\sqrt{1-x^2}]
(c) 3\sin^{-1} x = \sin^{-1}(3x - 4x^3)
Proof: Let \sin^{-1} x = \theta \Rightarrow \sin\theta = x
We know that \sin 3\theta = 3\sin\theta - 4\sin^3\theta
\Rightarrow \sin 3\theta = 3x - 4x^3
\Rightarrow 3\theta = \sin^{-1}(3x - 4x^3)
\Rightarrow 3\sin^{-1} x = \sin^{-1}(3x - 4x^3)
(d) \cos^{-1} x \pm \cos^{-1} y = \cos^{-1}\leqft[xy \mp \sqrt{1-x^2}\sqrt{1-y^2}\right]
Proof: Let \cos^{-1} x = \theta_1 \text{ i.e. } \cos\theta_1 = x
\cos^{-1} y = \theta_2 \text{ i.e. } \cos\theta_2 = y
\sin\theta_1 = \sqrt{1-\cos^2\theta_1} = \sqrt{1-x^2}
\sin\theta_2 = \sqrt{1-\cos^2\theta_2} = \sqrt{1-y^2}
We know that
\cos(\theta_1 \pm \theta_2) = \cos\theta_1\cos\theta_2 \mp \sin\theta_1\sin\theta_2
= xy \mp \sqrt{1-x^2}\sqrt{1-y^2}
\Rightarrow \theta_1 \pm \theta_2 = \cos^{-1}\left(xy \mp \sqrt{1-x^2}\sqrt{1-y^2}\right)
\Rightarrow \cos^{-1} x \pm \cos^{-1} y = \cos^{-1}\left(xy \mp \sqrt{1-x^2}\sqrt{1-y^2}\right)
(e) 2\cos^{-1} x = \cos^{-1}(2x^2 - 1)
Proof: Let \cos^{-1} x = \theta \Rightarrow \cos\theta = x
\cos 2\theta = 2\cos^2\theta - 1
\Rightarrow 2\theta = \cos^{-1}(2\cos^2\theta - 1)
\Rightarrow 2\cos^{-1} x = \cos^{-1}(2x^2 - 1)
(f) 3\cos^{-1} x = \cos^{-1}(4x^3 - 3x)
Proof: Let \cos^{-1} x = \theta \Rightarrow \cos\theta = x
\cos 3\theta = 4\cos^3\theta - 3\cos\theta
\Rightarrow 3\theta = \cos^{-1}(4\cos^3\theta - 3\cos\theta)
= \cos^{-1}(4x^3 - 3x)
\Rightarrow 3\cos^{-1} x = \cos^{-1}(4x^3 - 3x)
(5.) Prove that
(a) \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)
Proof: Let \tan^{-1} x = \theta_1 \Rightarrow \tan\theta_1 = x
and \tan^{-1} y = \theta_2 \Rightarrow \tan\theta_2 = y
\tan(\theta_1 + \theta_2) = \frac{\tan\theta_1 + \tan\theta_2}{1 - \tan\theta_1 \tan\theta_2}
\Rightarrow \theta_1 + \theta_2 = \tan^{-1}\left(\frac{\tan\theta_1 + \tan\theta_2}{1 - \tan\theta_1 \tan\theta_2}\right)
\Rightarrow \theta_1 + \theta_2 = \tan^{-1}\left(\frac{x+y}{1-xy}\right)
\Rightarrow \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)
(b) \tan^{-1} x - \tan^{-1} y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)
Proof: We can prove similarly 5(a).
(c) \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \tan^{-1}\left(\frac{x+y+z-xyz}{1-xy-yz-xz}\right)
Proof:
\tan^{-1} x + \tan^{-1} y + \tan^{-1} z
= \tan^{-1}\left(\frac{x+y}{1-xy}\right) + \tan^{-1} z \quad [\text{from } 5(a)]
= \tan^{-1}\left(\frac{\frac{x+y}{1-xy} + z}{1 - z \cdot \left(\frac{x+y}{1-xy}\right)}\right)
= \tan^{-1}\left(\frac{x+y+z-xyz}{1-xy-yz-xz}\right)
(d) 2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)
Proof: Let \tan^{-1} x = \theta \text{ i.e. } \tan\theta = x
\therefore \tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta} = \frac{2x}{1-x^2}
\Rightarrow 2\theta = \tan^{-1}\left(\frac{2x}{1-x^2}\right)
\Rightarrow 2\tan^{-1} x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)
(e) We know that
\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}
\Rightarrow 3\theta = \tan^{-1}\left(\frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}\right)
= \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)
\Rightarrow 3\tan^{-1} x = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right)
(6.) Prove that
(a) \cot^{-1} x + \cot^{-1} y = \cot^{-1}\left(\frac{xy - 1}{x + y}\right)
Proof: Let \cot^{-1} x = \theta_1 \text{ and } \cot^{-1} y = \theta_2
then \cot\theta_1 = x, \quad \cot\theta_2 = y
We know that
\cot(\theta_1 + \theta_2) = \frac{\cot\theta_1 \cot\theta_2 - 1}{\cot\theta_1 + \cot\theta_2}
\Rightarrow \theta_1 + \theta_2 = \cot^{-1}\left(\frac{\cot\theta_1 \cot\theta_2 - 1}{\cot\theta_1 + \cot\theta_2}\right)
\Rightarrow \cot^{-1} x + \cot^{-1} y = \cot^{-1}\left(\frac{xy - 1}{x + y}\right)
Similarly we can prove
\cot^{-1} x - \cot^{-1} y = \cot^{-1}\left(\frac{xy + 1}{y - x}\right)
(7.) Prove that
(a) \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}
Proof: Let \sin^{-1} x = \theta \text{ then } x = \sin\theta
\Rightarrow x = \cos\left(\frac{\pi}{2} - \theta\right)
\Rightarrow \cos^{-1} x = \frac{\pi}{2} - \theta
\Rightarrow \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x
\Rightarrow \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}
(b) \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}
Proof: \tan^{-1} x = \theta \Rightarrow x = \tan\theta
\Rightarrow x = \cot\left(\frac{\pi}{2} - \theta\right)
\Rightarrow \cot^{-1} x = \frac{\pi}{2} - \theta
\Rightarrow \cot^{-1} x = \frac{\pi}{2} - \tan^{-1} x
\Rightarrow \tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}
(c) \sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2}
Proof: Let \sec^{-1} x = \theta \Rightarrow x = \sec\theta
\Rightarrow x = \operatorname{cosec}\left(\frac{\pi}{2} - \theta\right)
\Rightarrow \operatorname{cosec}^{-1} x = \frac{\pi}{2} - \theta
\Rightarrow \operatorname{cosec}^{-1} x = \frac{\pi}{2} - \sec^{-1} x
\Rightarrow \sec^{-1} x + \operatorname{cosec}^{-1} x = \frac{\pi}{2}
By solving the above examples,one can understand Properties of Inverse Trig Functions.

Also Read This Article:- Types of Functions in Class 12

3.Practice Problems of Properties of Inverse Trig Functions for Students

(1.) Evaluate
\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right), (x < \pi)
(2.) Evaluate
\tan^{-1}\left(\frac{a - b}{1 + ab}\right) + \tan^{-1}\left(\frac{b - c}{1 + bc}\right) + \tan^{-1}\left(\frac{c - a}{1 + ca}\right)
Answers: (1.) \frac{\pi}{4} - x  (2.) 0
By solving the above questions,you can understand the Properties of Inverse Trig Functions well because the concept is well understood when you solve it practically.

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Also Read This Article:- Inverse trigonometric functions

4.Frequently Asked Questions Related to Properties of Inverse Trig Functions

Q:1.The inverse trigonometric function is not valid for which values?

Ans:It may be mentioned here that these results are valid within the principal value branches of the corresponding inverse trigonometric functions and wherever they are defined.Some results may not be valid for all values of the domains of inverse trigonometric functions.In fact,they will be valid only for some values of x for which inverse trigonometric functions.

Q:2.What is meant by inverse trigonometric function?

Ans:Inverse Trigonometric Functions are functions that perform the exact opposite (inverse) function of normal trigonometric functions.
Difference in Simple Terms:
General trigonometric function: Gives the ratio when given the value of the angle.
Inverse trigonometric function: Gives its original angle when the ratio is given.
Key Condition: The inverse of a trigonometric function is defined only in its Restricted Domain and Range, as normal trigonometric functions are periodic.

Q:3.Define mathematics according to FELIX KLEIN

Ans:Mathematics,in general,is fundamentally the science of self-evident things.
By answering the above questions,you can know about the primary terms of Properties of Inverse Trig Functions.

5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
\sqrt{900} + \sqrt{0.009} - \sqrt{0.000009}  का मान है:
(The value of \sqrt{900} + \sqrt{0.009} - \sqrt{0.000009}  is):
*पिछली प्रश्नोत्तरी का हल*
माना संख्या x है
\therefore x का \frac{1}{4} का \frac{2}{5} \times \frac{1}{3} = 40
\Rightarrow x \times \frac{1}{4} \times \frac{2}{5} \times \frac{1}{3} = 40
\Rightarrow x = \frac{40 \times 4 \times 5 \times 3}{2} = 1200
*Previous Quiz Solution*
Let the number be x
x of \frac{1}{3} of \frac{1}{4} \times \frac{2}{5} = 40
\Rightarrow x \times \frac{1}{3} \times \frac{1}{4} \times \frac{2}{5} = 40
\Rightarrow x = \frac{40 \times 3 \times 4 \times 5}{2}
\Rightarrow x = 1200
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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